Solution+Calculus+by+Howard+Anton+7th+ed+(Part+1) by Ali Ahmad

Bismillah hir Rehman nir Raheem ------------------------------------Assalat o Wasalam o Allika Ya RasoolALLAH

Calculus Solution Manuals(7

Ed.)

Anton, Bivens, Davis Ch Ch

Published By: Muhammad Hassan Riaz Yousufi

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Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH CHAPTER 1

EXERCISE SET 1.1

us uf

Functions

1. (a) around 1943 (b) 1960; 4200 (c) no; you need the year’s population (d) war; marketing techniques (e) news of health risk; social pressure, antismoking campaigns, increased taxation

2. (a) 1989; $35,600 (b) 1975, 1983; $32,000 (c) the ﬁrst two years; the curve is steeper (downhill)

(b) none (c) y = 0 (e) ymax = 2.8 at x = −2.6; ymin = −2.2 at x = 1.2

4. (a) x = −1, 4 (d) x = 0, 3, 5

(b) none (c) y = −1 (e) ymax = 9 at x = 6; ymin = −2 at x = 0 (b) none

6. (a) x = 9

(b) none

(d) ymin = −1; no maximum value

5. (a) x = 2, 4

3. (a) −2.9, −2.0, 2.35, 2.9 (d) −1.75 ≤ x ≤ 2.15

(d) ymin = 1; no maximum value

7. (a) Breaks could be caused by war, pestilence, ﬂood, earthquakes, for example. (b) C decreases for eight hours, takes a jump upwards, and then repeats.

8. (a) Yes, if the thermometer is not near a window or door or other source of sudden temperature change. (b) No; the number is always an integer, so the changes are in movements (jumps) of at least one unit.

(c)

9. (a) The side adjacent to the building has length x, so L = x + 2y. Since A = xy = 1000, L = x + 2000/x. (b) x > 0 and x must be smaller than the width of the building, which was not given.

120

(d) Lmin ≈ 89.44 ft

10. (a) V = lwh = (6 − 2x)(6 − 2x)x

(b) From the ﬁgure it is clear that 0 < x < 3.

(c)

uh a 0

(d) Vmax ≈ 16 in3

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Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Chapter 1

500 . Then πr2

C = (0.02)(2)πr2 + (0.01)2πrh = 0.04πr2 + 0.02πr = 0.04πr2 +

500 πr2

11. (a) V = 500 = πr2 h so h =

10 ; Cmin ≈ 4.39 at r ≈ 3.4, h ≈ 13.8. r

us uf

1.5

10 (b) C = (0.02)(2)(2r)2 + (0.01)2πrh = 0.16r2 + . Since r 0.04π < 0.16, the top and bottom now get more weight. Since they cost more, we diminish their sizes in the solution, and the cans become taller.

1.5

5.5

(b) P = 2L + 2πr = 1320 and 2r = 2x + 160, so L = 12 (1320 − 2πr) = 12 (1320 − 2π(80 + x)) = 660 − 80π − πx.

12. (a) The length of a track with straightaways of length L and semicircles of radius r is P = (2)L + (2)(πr) ft. Let L = 360 and r = 80 to get P = 720 + 160π = 1222.65 ft. Since this is less than 1320 ft (a quarter-mile), a solution is possible. 450

100 0

EXERCISE SET 1.2

(c) The shortest straightaway is L = 360, so x = 15.49 ft. (d) The longest straightaway occurs when x = 0, so L = 660 − 80π = 408.67 ft.

1. (a) f (0) = 3(0)2 −2 = −2; f (2) = 3(2)2 −2 = 10; f (−2) = 3(−2)2 −2 = 10; f (3) = 3(3)2 −2 = 25; √ √ f ( 2) = 3( 2)2 − 2 = 4; f (3t) = 3(3t)2 − 2 = 27t2 − 2 √ √ (b) f (0) = 2(0) = 0; f (2) = 2(2) = 4; f (−2) = 2(−2) = −4; f (3) = 2(3) = 6; f ( 2) = 2 2; f (3t) = 1/3t for t > 1 and f (3t) = 6t for t ≤ 1. −1 + 1 π+1 −1.1 + 1 −0.1 1 3+1 = 2; g(−1) = = 0; g(π) = ; g(−1.1) = = = ; 3−1 −1 − 1 π−1 −1.1 − 1 −2.1 21 2 2 t −1+1 t g(t2 − 1) = 2 = 2 t −1−1 t −2 √ √ (b) g(3) = 3 + 1 = 2; g(−1) = 3; g(π) = π + 1; g(−1.1) = 3; g(t2 − 1) = 3 if t2 < 2 and √ g(t2 − 1) = t2 − 1 + 1 = |t| if t2 ≥ 2.

uh a

2. (a) g(3) =

√ √ 3. (a) x = 3 (b) x ≤ − 3 or x ≥ 3 (c) x2 − 2x + 5 = 0 has no real solutions so x2 − 2x + 5 is always positive or always negative. If x = 0, then x2 − 2x + 5 = 5 > 0; domain: (−∞, +∞). (d) x = 0 (e) sin x = 1, so x = (2n + 12 )π, n = 0, ±1, ±2, . . .

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Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Exercise Set 1.2

4. (a) x = − 75

(b) −2 ≤ x ≤ 2

(d) all x

6. (a) x ≥

(b) − 32 ≤ x ≤

(d) x = 0

2 3

3 2

7. (a) yes (c) no (vertical line test fails)

(e) all x

5. (a) x ≤ 3

us uf

(b) x − 3x2 must be nonnegative; y = x − 3x2 is a parabola that crosses the x-axis at x = 0, 13 and opens downward, thus 0 ≤ x ≤ 13 x2 − 4 > 0, so x2 − 4 > 0 and x − 4 > 0, thus x > 4; or x2 − 4 < 0 and x − 4 < 0, thus (c) x−4 −2 < x < 2 (d) x = −1 (e) cos x ≤ 1 < 2, 2 − cos x > 0, all x

(e) x ≥ 0

(b) yes (d) no (vertical line test fails)

8. The sine of θ/2 is (L/2)/10 (side opposite over hypotenuse), so that L = 20 sin(θ/2).

11.

t 5

12.

10.

9. The cosine of θ is (L − h)/L (side adjacent over hypotenuse), so h = L(1 − cos θ).

13. (a) If x < 0, then |x| = −x so f (x) = −x + 3x + 1 = 2x + 1. If x ≥ 0, then |x| = x so f (x) = x + 3x + 1 = 4x + 1; 2x + 1, x < 0 f (x) = 4x + 1, x ≥ 0

(b) If x < 0, then |x| = −x and |x − 1| = 1 − x so g(x) = −x + 1 − x = 1 − 2x. If 0 ≤ x < 1, then |x| = x and |x − 1| = 1 − x so g(x) = x + 1 − x = 1. If x ≥ 1, then |x| = x and |x − 1| = x − 1 so g(x) = x + x − 1 = 2x − 1;  x<0  1 − 2x, 1, 0≤x<1 g(x) =  2x − 1, x≥1

uh a

14. (a) If x < 5/2, then |2x − 5| = 5 − 2x so f (x) = 3 + (5 − 2x) = 8 − 2x. If x ≥ 5/2, then |2x − 5| = 2x − 5 so f (x) = 3 + (2x − 5) = 2x − 2; 8 − 2x, x < 5/2 f (x) = 2x − 2, x ≥ 5/2 (b) If x < −1, then |x − 2| = 2 − x and |x + 1| = −x − 1 so g(x) = 3(2 − x) − (−x − 1) = 7 − 2x. If −1 ≤ x < 2, then |x − 2| = 2 − x and |x + 1| = x + 1 so g(x) = 3(2 − x) − (x + 1) = 5 − 4x. If x ≥ 2, then |x − 2| = x − 2 and |x + 1| = x + 1 so g(x) = 3(x − 2) − (x + 1) = 2x − 7;  x < −1  7 − 2x, 5 − 4x, −1 ≤ x < 2 g(x) =  2x − 7, x≥2

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Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Chapter 1

us uf

15. (a) V = (8 − 2x)(15 − 2x)x (b) −∞ < x < +∞, −∞ < V < +∞ (c) 0 < x < 4 (d) minimum value at x = 0 or at x = 4; maximum value somewhere in between (can be approximated by zooming with graphing calculator) (b) θ = nπ + π/2 for n an integer, −∞ < n < ∞ (d) 3000 ft

16. (a) x = 3000 tan θ (c) 0 ≤ θ < π/2, 0 ≤ x < +∞

6000

0 0

x = 1, −2 causes division by zero

(ii)

g(x) = x + 1, all x

18. (i)

x = 0 causes division by zero

(ii)

g(x) =

√

x + 1 for x ≥ 0

19. (a) 25◦ F

17. (i)

(b) 2◦ F

20. If v = 48 then −60 = WCI = 1.6T − 55; thus T = (−60 + 55)/1.6 ≈ −3◦ F.

√ 21. If v = 8 then −10 = WCI = 91.4 + (91.4 −√T )(0.0203(8) − 0.304 8 − 0.474); thus T = 91.4 + (10 + 91.4)/(0.0203(8) − 0.304 8 − 0.474) and T = 5◦ F

22. The WCI is given by three formulae, but the ﬁrst√and third don’t work√with the data. Hence −15 = WCI = 91.4 + (91.4 − 20)(0.0203v − 0.304 v − 0.474); set x = v so that v = x2 and obtain 0.0203x2 − 0.304x − 0.474 + (15 + 91.4)/(91.4 − 20) = 0. Use the quadratic formula to ﬁnd the two roots. Square them to get v and discard the spurious solution, leaving v ≈ 25. 23. Let t denote time in minutes after 9:23 AM. Then D(t) = 1000 − 20t ft.

EXERCISE SET 1.3

-0.5

2. (e) seems best, though only (a) is bad and (b) is not good.

3. (b) and (c) are good; (a) is very bad. y

y 0.5

uh a

-1

0.5

1. (e) seems best, though only (a) is bad.

x -1

-1

x 1

-0.5

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Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Exercise Set 1.3

5. [−3, 3] × [0, 5]

4. (b) and (c) are good; (a) is very bad.

6. [−4, 2] × [0, 3]

us uf

-11 -12

-13

–2

x 2

-14 x -1

-2

7. (a) window too narrow, too short (c) good window, good spacing

(b) window wide enough, but too short (d) window too narrow, too short

y x 5

-5 -200

-400

(e) window too narrow, too short

(b) window too short (d) window too narrow, too short

8. (a) window too narrow (c) good window, good tick spacing y 50 -8 -4

-16

-100

-250

(e) shows one local minimum only, window too narrow, too short 9. [−5, 14] × [−60, 40]

10. [6, 12] × [−100, 100]

y 40 20

y 100

3 2

-0.1

0.1

-50

-5

11. [−0.1, 0.1] × [−3, 3]

-2

-60

uh a

12. [−1000, 1000] × [−13, 13] y

y -1000

13. [−250, 1050] × [−1500000, 600000]

-1000

x 1000

-500000

1000

-5

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Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Chapter 1

14. [−3, 20] × [−3500, 3000]

15. [−2, 2] × [−20, 20]

16. [1.6, 2] × [0, 2]

y 2

-2

-1000

-1

2 0.5

-10

us uf

x 5

1000

-2000

-20

4 2

-4 -2

√

-2

19. (a) f (x) =

-4

√ (b) f (x) = − 16 − x2

16 − x2 y

x 2 y

2 x -2

-2

(c)

18. depends on graphing utility

17. depends on graphing utility

1.8

1.6 1.7

(d)

-4

-2

-1

-3

1 x

-2

(e) No; the vertical line test fails.

√ (b) y = ± x2 + 1

1 − x2 /4

20. (a) y = ±3

y 4

uh a

x x

-2

-4

-2

-2 -4

-3

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Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Exercise Set 1.3

21. (a)

(b) 1

x -1

1 y

(c)

(d)

x -1

-1

–1

22.

(f ) c

-1

(e)

us uf

-1

x 1

-1

23. The portions of the graph of y = f (x) which lie below the x-axis are reﬂected over the x-axis to give the graph of y = |f (x)|.

24. Erase the portion of the graph of y = f (x) which lies in the left-half plane and replace it with the reﬂection over the y-axis of the portion in the right-half plane (symmetry over the y-axis) and you obtain the graph of y = f (|x|). 25. (a) for example, let a = 1.1

uh a

3 2 1 x x

a 1

26. They are identical.

(b)

27. 15

10 5 x

x -1

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Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Chapter 1

28. This graph is very complex. We show three views, small (near the origin), medium and large: y

(a)

(b)

(c)

-50

-1

1000

-1

us uf

10 x

29. (a)

(b) 1 0.5

x -1

(c)

(d)

1.5

0.5 1

-1

0.5 x -1

-2

30.

-3

-1 -0.5

-3

1 x

31. (a) stretches or shrinks the graph in the y-direction; reﬂects it over the x-axis if c changes sign y

uh a

c=2

c=1

-2 -1

x c = –1.5 x 2

-2

c = –1.5

c=2 c = 0.5 c = -1

c=1

-1

(b) As c increases, the parabola moves down and to the left. If c increases, up and right.

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Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Exercise Set 1.4

32. (a)

us uf

x 1

–2

(b) x-intercepts at x = 0, a, b. Assume a < b and let a approach b. The two branches of the curve come together. If a moves past b then a and b switch roles. y

1 -1

-1

a = 0.5 b = 1.5

-3

a=1 b = 1.5

-1 a = 1.5 b = 1.6

-3

34. The curve oscillates between the lines y = +1 and y = −1, inﬁnitely many times in any neighborhood of x = 0.

-10 -10

x 10

EXERCISE SET 1.4 y

(b)

2 x 1

-1

uh a

(c)

2 y

(d)

2 x 1

x -4

-2

-1

–1

-1

-30

1. (a)

x –2

-30

30 10

x 2

33. The curve oscillates between the lines y = x and y = −x with increasing rapidity as |x| increases.

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Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Chapter 1

2. (a)

y 1

us uf

-3

(b)

x -1 y

(d)

(c)

x 1

-1

-1 y

(b)

3. (a)

1 x

-1

-2

-1

1 y x -1

y 1 x

-1

(d)

1 x

-1

(c)

-1

5. Translate right 2 units, and up one unit. y

6. Translate left 1 unit, reﬂect over x-axis, and translate up 2 units.

uh a

x -2

7. Translate left 1 unit, stretch vertically by a factor of 2, reﬂect over x-axis, translate down 3 units. y -6

-2 -20

x 2

1 -60 –2

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Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Exercise Set 1.4

10. y = (x + 3)2 − 19; translate left 3 units and down 19 units.

–4

9. y = (x + 3)2 − 9; translate left 3 units and down 9 units.

us uf

8. Translate right 3 units, compress vertically by a factor of 12 , and translate up 2 units.

–5

x -8

-4 -2

2 -5

x 4

12. y = 12 [(x − 1)2 + 2]; translate left 1 unit and up 2 units, compress vertically by a factor of 12 .

x -2 -1 -2

13. Translate left 1 unit, reﬂect over x-axis, translate up 3 units.

11. y = −(x − 1)2 + 2; translate right 1 unit, reﬂect over x-axis, translate up 2 units.

3 2

x 2

1 -6

15. Compress vertically by a factor of 12 , translate up 1 unit.

16. Stretch vertically by √ a factor of 3 and reﬂect over x-axis.

14. Translate right 4 units and up 1 unit. y

2 -1

17. Translate right 3 units. y

x 1

uh a M

18. Translate right 1 unit and reﬂect over x-axis.

-10

19. Translate left 1 unit, reﬂect over x-axis, translate up 2 units. y

x 4

6 -2

2 -4

2 -3

-1-2

x 1

-6

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Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Chapter 1

20. y = 1 − 1/x; reﬂect over x-axis, translate up 1 unit.

21. Translate left 2 units and down 2 units.

y x

5 -4

-2

x 2

-2

22. Translate right 3 units, reﬂect over x-axis, translate up 1 unit.

-5

us uf

23. Stretch vertically by a factor of 2, translate right 1 unit and up 1 unit.

y 4

x 5 -1

24. y = |x − 2|; translate right 2 units. 2

1 x 4

-1

28. Translate right 2 units, reﬂect over x-axis.

1 x

uh a 29. (a)

4 1 x

–2

x -2

27. Translate left 1 unit and up 2 units.

26. Translate right 2 units and down 3 units.

25. Stretch vertically by a factor of 2, reﬂect over x-axis, translate up 1 unit.

-3

-1

–1

(b) y =

0 if x ≤ 0 2x if 0 < x

x -1

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Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Exercise Set 1.4

30.

us uf

x 2 –5

31. (f + g)(x) = x2 + 2x + 1, all x; (f − g)(x) = 2x − x2 − 1, all x; (f g)(x) = 2x3 + 2x, all x; (f /g)(x) = 2x/(x2 + 1), all x

32. (f + g)(x) = 3x − 2 + |x|, all x; (f − g)(x) = 3x − 2 − |x|, all x; (f g)(x) = 3x|x| − 2|x|, all x; (f /g)(x) = (3x − 2)/|x|, all x = 0

√ √ 33. (f + g)(x) = 3 x − 1, x ≥ 1; (f − g)(x) = x − 1, x ≥ 1; (f g)(x) = 2x − 2, x ≥ 1; (f /g)(x) = 2, x > 1

34. (f + g)(x) = (2x2 + 1)/[x(x2 + 1)], all x = 0; (f − g)(x) = −1/[x(x2 + 1)], all x = 0; (f g)(x) = 1/(x2 + 1), all x = 0; (f /g)(x) = x2 /(x2 + 1), all x = 0 (b) 9

(d) 2

36. (a) π − 1

(b) 0

(d) 1

37. (a) t4 + 1

(b) t2 + 4t + 5

√

5s + 2 √ (e) 4 x

(b)

√

(f ) 0

(g) x + 1

x+2

√ (c) 3 5x √ (g) 1/ 4 x

1 +1 x2 (h) 9x2 + 1 (d)

√ (d) 1/ x (h) |x − 1|

38. (a)

(f ) x2 + 1

(e) x2 + 2xh + h2 + 1

35. (a) 3

39. (f ◦ g)(x) = 2x2 − 2x + 1, all x; (g ◦ f )(x) = 4x2 + 2x, all x 40. (f ◦ g)(x) = 2 − x6 , all x; (g ◦ f )(x) = −x6 + 6x4 − 12x2 + 8, all x

42. (f ◦ g)(x) =

41. (f ◦ g)(x) = 1 − x, x ≤ 1; (g ◦ f )(x) = √

x2 + 3 − 3, |x| ≥

√

1 − x2 , |x| ≤ 1

6; (g ◦ f )(x) =

√

x, x ≥ 3

1 1 1 1 , x = , 1; (g ◦ f )(x) = − − , x = 0, 1 1 − 2x 2 2x 2

44. (f ◦ g)(x) =

x 1 , x = 0; (g ◦ f )(x) = + x, x = 0 x2 + 1 x

43. (f ◦ g)(x) =

uh a

45. x−6 + 1

47. (a) g(x) =

√

46. x, h(x) = x + 2

48. (a) g(x) = x + 1, h(x) = x2

49. (a) g(x) = x2 , h(x) = sin x

x x+1 (b) g(x) = |x|, h(x) = x2 − 3x + 5 (b) g(x) = 1/x, h(x) = x − 3 (b) g(x) = 3/x, h(x) = 5 + cos x

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Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Chapter 1

50. (a) g(x) = 3 sin x, h(x) = x2

(b) g(x) = 3x2 + 4x, h(x) = sin x

51. (a) f (x) = x3 , g(x) = 1 + sin x, h(x) = x2

(b) f (x) =

52. (a) f (x) = 1/x, g(x) = 1 − x, h(x) = x2

(b) f (x) = |x|, g(x) = 5 + x, h(x) = 2x

2 1

x, g(x) = 1 − x, h(x) =

54. {−2, −1, 0, 1, 2, 3}

y x 1

-3 -2 -1 -1 -2 -3 -4

55. Note that f (g(−x)) = f (−g(x)) = f (g(x)), so f (g(x)) is even.

56. Note that g(f (−x)) = g(f (x)), so g(f (x)) is even. y

f (g(x))

3 g( f (x))

–1 –1

–3

–1 –1

–3

√ 3

–2

–3

57. f (g(x)) = 0 when g(x) = ±2, so x = ±1.4; g(f (x)) = 0 when f (x) = 0, so x = ±2. 58. f (g(x)) = 0 at x = −1 and g(f (x)) = 0 at x = −1 59.

6xh + 3h2 3(x + h)2 − 5 − (3x2 − 5) = 6x + 3h; = h h

3w2 − 5 − (3x2 − 5) 3(w − x)(w + x) = 3w + 3x = w−x w−x (x + h)2 + 6(x + h) − (x2 + 6x) 2xh + h2 + 6h = = 2x + h + 6; h h

60.

w2 + 6w − (x2 + 6x) =w+x+6 w−x x − (x + h) −1 1/w − 1/x x−w 1 1/(x + h) − 1/x = = ; = =− h xh(x + h) x(x + h) w−x wx(w − x) xw

62.

x2 − (x + h)2 x2 − w2 2x + h 1/w2 − 1/x2 1/(x + h)2 − 1/x2 x+w = 2 = = − ; =− 2 2 2 2 2 2 2 h x h(x + h) x (x + h) w−x x w (w − x) x w

uh a

61.

63. (a) the origin

(b) the x-axis

us uf

53.

√

(d) none

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Exercise Set 1.4

64. (a)

(b)

us uf

(c)

−3 1

−2 −5

−1 −1

0 0

1 −1

(b)

x f (x)

−3 1

−2 5

−1 −1

0 0

1 1

2 −5 2 −5

3 1 3 −1

66. (a)

x f (x)

65. (a)

(b)

68. neither; odd; even

(b) odd

67. (a) even

69. (a) f (−x) = (−x)2 = x2 = f (x), even (c) f (−x) = | − x| = |x| = f (x), even

(f )

(d) neither

(b) f (−x) = (−x)3 = −x3 = −f (x), odd (d) f (−x) = −x + 1, neither

x3 + x (−x)3 − (−x) = − = −f (x), odd 1 + (−x)2 1 + x2

(e) f (−x) =

f (−x) = 2 = f (x), even

70. (a) x-axis, because x = 5(−y)2 + 9 gives x = 5y 2 + 9

(b) x-axis, y-axis, and origin, because x2 − 2(−y)2 = 3, (−x)2 − 2y 2 = 3, and (−x)2 − 2(−y)2 = 3 all give x2 − 2y 2 = 3

uh a

(b) origin, because (−y) =

x (−x) gives y = 3 + (−x)2 3 + x2

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Chapter 1

73.

-4

-3

–3

-2

72.

us uf

74. (a) Whether we replace x with −x, y with −y, or both, we obtain the same equation, so by Theorem 1.4.3 the graph is symmetric about the x-axis, the y-axis and the origin. (b) y = (1 − x2/3 )3/2

75.

76.

2 y

77. (a)

78. (a)

(b)

(c) For quadrant II, the same; for III and IV use y = −(1 − x2/3 )3/2 . (For graphing it may be helpful to use the tricks that precede Exercise 29 in Section 1.3.)

2‚ 1 ‚ 3

-‚ 2 -1

(d)

y 1 x

uh a

-1

(c)

(b)

-1

c/2

x 1

79. Yes, e.g. f (x) = xk and g(x) = xn where k and n are integers.

80. If x ≥ 0 then |x| = x and f (x) = g(x). If x < 0 then f (x) = |x|p/q if p is even and f (x) = −|x|p/q if p is odd; in both cases f (x) agrees with g(x).

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Exercise Set 1.5

EXERCISE SET 1.5 3−0 3 3 − (8/3) 1 0 − (8/3) 2 =− , =− , = 0−2 2 0−6 18 2−6 3 (b) Yes; the ﬁrst and third slopes above are negative reciprocals of each other. −1 − (−1) −1 − 3 3−3 −1 − 3 = 0, = 2, = 0, =2 −3 − 5 5−7 7 − (−1) −3 − (−1)

3. III < II < IV < I

4. III < IV < I < II

(b) Yes; there are two pairs of equal slopes, so two pairs of parallel lines.

us uf

2. (a)

1. (a)

1 − (−5) −5 − (−1) 1 − (−1) = 2. Since the slopes connecting all pairs of points = 2, = 2, 1 − (−2) −2 − 0 1−0 are equal, they lie on a line. 2−5 4−5 1 4−2 = −1, = 3, = . Since the slopes connecting the pairs of points are (b) −2 − 0 0−1 −2 − 1 3 not equal, the points do not lie on a line.

(a) If x = 9 then y = 1.

y−2 , and thus y − 2 = 3(x − 1). x−1 (b) If y = −2 then x = −1/3.

7. The slope, m = 3, is equal to

y−5 , and thus y − 5 = −2(x − 7). x−7 (b) If y = 12 then x = 7/2.

6. The slope, m = −2, is obtained from

5. (a)

(a) If x = 5 then y = 14.

y−0 1 y−5 = or y = x/2. Also = 2 or y = 2x − 9. Solve x−0 2 x−7 simultaneously to obtain x = 6, y = 3.

8. (a) Compute the slopes:

5−0 2−0 and the second is . Since they are negatives of each other we 1−x 4−x get 2(4 − x) = −5(1 − x) or 7x = 13, x = 13/7.

9. (a) The ﬁrst slope is

(b) 135◦

(d) 91◦

11. (a) 153◦

(b) 45◦

(d) 89◦

10. (a) 27◦

√ 12. (a) m = tan φ = − 3/3, so φ = 150◦ √

3, so φ = 60◦

13. (a) m = tan φ =

(b) m = tan φ = 4, so φ = 76◦ (b) m = tan φ = −2, so φ = 117◦

14. y = 0 and x = 0 respectively 15. y = −2x + 4

16. y = 5x − 3 2

uh a

6 –1

-1

1 0

–8

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Chapter 1

18. The slope of both lines is −3/2, so y − 2 = (−3/2)(x − (−1)), or y = − 32 x + 12

17. Parallel means the lines have equal slopes, so y = 4x + 7.

-2 -1

1 –4

20. The slope of x − 4y = 7 is 1/4 whose negative reciprocal is −4, so y − (−4) = −4(x − 3) or y = −4x + 8.

19. The negative reciprocal of 5 is −1/5, so y = − 15 x + 6.

-9

us uf

–3

4 − (4 − 7) = 11, 2−1 so y − (−7) = 11(x − 1), or y = 11x − 18

ss Ha

6−1 = −5, so −3 − (−2) y − 6 = −5(x − (−3)), or y = −5x − 9

22. m =

21. m =

–3

0 –5

-9

23. (a) m1 = m2 = 4, parallel (b) m1 = 2 = −1/m2 , perpendicular (c) m1 = m2 = 5/3, parallel (d) If A = 0 and B = 0 then m1 = −A/B = −1/m2 , perpendicular; if A = 0 or B = 0 (not both) then one line is horizontal, the other vertical, so perpendicular. (e) neither

uh a

24. (a) m1 = m2 = −5, parallel (c) m1 = −4/5 = −1/m2 , perpendicular (e) neither

(b) m1 = 2 = −1/m2 , perpendicular (d) If B = 0, m1 = m2 = −A/B; if B = 0 both are vertical, so parallel

25. (a) m = (0 − (−3))/(2 − 0)) = 3/2 so y = 3x/2 − 3 (b) m = (−3 − 0)/(4 − 0) = −3/4 so y = −3x/4

26. (a) m = (0 − 2)/(2 − 0)) = −1 so y = −x + 2 (b) m = (2 − 0)/(3 − 0) = 2/3 so y = 2x/3

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Exercise Set 1.5

27. (a) The velocity is the slope, which is

5 − (−4) = 9/10 ft/s. 10 − 0

5−1 = 2 m/s 4−2

28. (a) v =

(b) x − 1 = 2(t − 2) or x = 2t − 3

13 3

30. (a) The acceleration is the slope of the velocity, so a = (c) v = 4 ft/s

5 1 0−5 =− = − ft/s2 . 10 − 0 10 2 (d) t = 4 s

(b) v = 5 ft/s

(b) v − 3 = − 43 (t − 1), or v = − 43 t +

3 − (−1) 4 = − ft/s2 . 1−4 3 13 (c) v = 3 ft/s

29. (a) The acceleration is the slope of the velocity, so a =

us uf

(b) x = −4 (c) The line has slope 9/10 and passes through (0, −4), so has equation x = 9t/10 − 4; at t = 2, x = −2.2. (d) t = 80/9

31. (a) It moves (to the left) 6 units with velocity v = −3 cm/s, then remains motionless for 5 s, then moves 3 units to the left with velocity v = −1 cm/s. 9 0−9 (b) vave = =− cm/s 10 − 0 10 (c) Since the motion is in one direction only, the speed is the negative of the velocity, so 9 save = 10 cm/s. 32. It moves right with constant velocity v = 5 km/h; then accelerates; then moves with constant, though increased, velocity again; then slows down.

33. (a) If x1 denotes the ﬁnal position and x0 the initial position, then v = (x1 − x0 )/(t1 − t0 ) = 0 mi/h, since x1 = x0 .

(b) If the distance traveled in one direction is d, then the outward journey took t = d/40 h. Thus 2d 80t total dist = = = 48 mi/h. save = total time t + (2/3)t t + (2/3)t (c) t + (2/3)t = 5, so t = 3 and 2d = 80t = 240 mi round trip

  if 0 ≤ t ≤ 10  10t 100 if 10 ≤ t ≤ 100 (b) v =   600 − 5t if 100 ≤ t ≤ 120

35. (a)

(b) v = 0 at t = 2

34. (a) down, since v < 0

t 80

uh a

36.

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Chapter 1

(b)

us uf

0.6

0.2 x 2

38. (a) Since y = 0.2 = (1)k, k = 1/5 and y = x/5 y

(b)

37. (a) y = 20 − 15 = 5 when x = 45, so 5 = 45k, k = 1/9, y = x/9

x 2

39. Each increment of 1 in the value of x yields the increment of 1.2 for y, so the relationship is linear. If y = mx + b then m = 1.2; from x = 0, y = 2, follows b = 2, so y = 1.2x + 2

40. Each increment of 1 in the value of x yields the increment of −2.1 for y, so the relationship is linear. If y = mx + b then m = −2.1; from x = 0, y = 10.5 follows b = 10.5, so y = −2.1x + 10.5 TC − 100 5 0 − 100 = , so TC = (TF − 32). TF − 212 32 − 212 9

41. (a) With TF as independent variable, we have (b) 5/9

42. (a) One degree Celsius is one degree Kelvin, so the slope is the ratio 1/1 = 1. Thus TC = TK − 273.15.

(b) TC = 0 − 273.15 = −273.15◦ C 43. (a)

p−1 5.9 − 1 = , or p = 0.098h + 1 h−0 50 − 0

44. (a)

133.9 − 123.4 R − 123.4 = , so R = 0.42T + 115. T − 20 45 − 20

r − 0.80 0.75 − 0.80 = , so r = −0.0125t + 0.8 t−0 4−0

(b) T = 32.38◦ C

(b) 64 days

uh a

45. (a)

(b) when p = 2, or h = 1/0.098 ≈ 10.20 m

46. (a) Let the position at rest be y0 . Then y0 + y = y0 + kx; with x = 11 we get y0 + kx = y0 + 11k = 40, and with x = 24 we get y0 + kx = y0 + 24k = 60. Solve to get k = 20/13 and y0 = 300/13.

(b) 300/13 + (20/13)W = 30, so W = (390 − 300)/20 = 9/2 g.

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Exercise Set 1.6

47. (a) For x trips we have C1 = 2x and C2 = 25 + x/4

(b) 2x = 25 + x/4, or x = 100/7, so the commuter pass becomes worthwhile at x = 15.

us uf

20 x 20

EXERCISE SET 1.6 1. (a) y = 3x + b

(b) y = 3x + 6

(c)

48. If the student drives x miles, then the total costs would be CA = 4000 + (1.25/20)x and CB = 5500 + (1.25/30)x. Set 4000 + 5x/80 = 5500 + 5x/120 and solve for x = 72, 000 mi.

y = 3x + 6

y = 3x + 2 y = 3x - 4 x

-2

-10

2. Since the slopes are negative reciprocals, y = − 13 x + b. 3. (a) y = mx + 2

(b) m = tan φ = tan 135◦ = −1, so y = −x + 2

(c)

m = 1.5 m=1

5 4 3

m = –1

-2

(b) y = m(x − 1) (d) 2x + 4y = C

5. (a) The slope is −1.

(b) The y-intercept is y = −1.

4. (a) y = mx (c) y = −2 + m(x − 1)

uh a

2 x

-1

-4

-2 -3

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Chapter 1

(d) The x-intercept is x = 1.

y 3

us uf

2 1

x -4

-6

-1

-2 -2

-3

(b) The y-intercept is y = −1/2.

6. (a) horizontal lines y

y 2

x 2

-1

x 1

(d) They pass through (−1, 1). y (-1, 1)

1 x -2

7. Let the line be tangent to the circle at the point (x0 , y0 ) where x20 + y02 = 9. The slope of the tangent line is the negative reciprocal of y0 /x0 (why?), so m = −x0 /y0 and y = −(x0 /y0 )x + b. 9 − x0 x Substituting the point (x0 , y0 ) as well as y0 = ± 9 − x20 we get y = ± . 9 − x20 1 3

8. Solve the simultaneous equations to get the point (−2, 1/3) of intersection. Then y =

+ m(x + 2).

9. The x-intercept is x = 10 so that with depreciation at 10% per year the ﬁnal value is always zero, and hence y = m(x − 10). The y-intercept is the original value.

uh a

x 2

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Exercise Set 1.6

(b) IV

(d) V

(e) I

(f ) II

us uf

11. (a) VI

10. A line through (6, −1) has the form y + 1 = m(x − 6). The intercepts are x = 6 + 1/m and y = −6m − 1. Set −(6 + 1/m)(6m + 1) = 3, or 36m2 + 15m + 1 = (12m + 1)(3m + 1) = 0 with roots m = −1/12, −1/3; thus y + 1 = −(1/3)(x − 6) and y + 1 = −(1/12)(x − 6).

13. (a)

-2

x 1

-10

12. In all cases k must be positive, or negative values would appear in the chart. Only kx−3 decreases, so that must be f (x). Next, kx2 grows faster than kx3/2 , so that would be g(x), which grows faster than h(x) (to see this, consider ratios of successive values of the functions). Finally, experimentation (a spreadsheet is handy) for values of k yields (approximately) f (x) = 10x−3 , g(x) = x2 /2, h(x) = 2x1.5 .

10 1

-40

-10

-30

(b)

x -2

2 x 2

-4

-2

-4

(c)

2 1

x -4

y 1 x

-1

-2

14. (a)

y 80

uh a

40 40 x

-2

x -2

-40

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Chapter 1

(b)

4 -2

us uf

-2 x -2

-4

(c)

-2

4 -2

x -1

-2

-3

-4

15. (a)

(b)

10 5

x -2

2 -5

-2

-10

(c)

10 5

-2 -5

16. (a)

-10

(b)

(c)

2 x

x 2

2 x 2

uh a

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Exercise Set 1.6

17. (a)

(b)

20 1

us uf

-1 -20

x 1

-1

-80

(c)

(d) x

-3

-10

-3

-20 -40

18. (a)

-40

(b)

1 x 1

-6

y 2 1 x

-4

-2

-1

4 x -2

2 -4

19. (a)

1.5

0.5

-3

-2

4 x -6

(b)

x 1

-1

uh a

(d)

30 10

x 3

-2

x 1

-20

-0.5

-1

-2

0.5

-1

(c)

(d)

(c)

-2

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Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Chapter 1

(b)

x 1

-10

-4 -2

-4

(d)

106

(c)

10 x -1

20. (a)

us uf

-106

21. y = x2 + 2x = (x + 1)2 − 1

-2

10 5

x -2

x 2

(b) k = 20 N·m

V (L)

0.25

1.0

0.5

40 × 10

20 × 10

1.5 3

13.3 × 10

2.0 3

10 × 103

P (N/m ) 80 × 10 (d)

-2

23. (a) N·m (c)

(b)

22. (a)

-2

P(N/m2)

30 20

uh a

V(m3) 10

24. If the side of the square base is x and the height of the container is y then V = x2 y = 100; minimize A = 2x2 + 4xy = 2x2 + 400/x. A graphing utility with a zoom feature suggests that the solution 1 is a cube of side 100 3 cm.

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Exercise Set 1.6

25. (a) F = k/x2 so 0.0005 = k/(0.3)2 and k = 0.000045 N·m2 . (b) F = 0.000005 N F (c)

us uf

10-5

d 10

(d) When they approach one another, the force becomes inﬁnite; when they get far apart it tends to zero. 26. (a) 2000 = C/(4000)2 , so C = 3.2 × 1010 lb·mi2 (b) W = C/50002 = (3.2 × 1010 )/(25 × 106 ) = 1280 lb. (d)

15000

5000 x 8000

2000

No, but W is very small when x is large.

(c)

(b) I; y = 0, x = −2, 3 (d) III; y = 0, x = −2

27. (a) II; y = 1, x = −1, 2 (c) IV; y = 2

28. The denominator has roots x = ±1, so x2 − 1 is the denominator. To determine k use the point (0, −1) to get k = 1, y = 1/(x2 − 1). 29. Order the six trigonometric functions as sin, cos, tan, cot, sec, csc: (a) pos, pos, pos, pos, pos, pos (b) neg, zero, undef, zero, undef, neg (c) pos, neg, neg, neg, neg, pos (d) neg, pos, neg, neg, pos, neg (e) neg, neg, pos, pos, neg, neg (f ) neg, pos, neg, neg, pos, neg

30. (a) neg, zero, undef, zero, undef, neg (c) zero, neg, zero, undef, neg, undef (e) neg, neg, pos, pos, neg, neg

31. (a) sin(π − x) = sin x; 0.588 (c) sin(2π + x) = sin x; 0.588 (e) cos2 x = 1 − sin2 x; 0.655

(b) pos, neg, neg, neg, neg, pos (d) pos, zero, undef, zero, undef, pos (f ) neg, neg, pos, pos, neg, neg (b) cos(−x) = cos x; 0.924 (d) cos(π − x) = − cos x; −0.924 (f ) sin2 2x = 4 sin2 x cos2 x = 4 sin2 x(1 − sin2 x); 0.905

uh a

32. (a) sin(3π + x) = − sin x; −0.588 (b) cos(−x − 2π) = cos x; 0.924 (c) sin(8π + x) = sin x; 0.588 (d) sin(x/2) = ± (1 − cos x)/2; use the negative sign for x small and negative; −0.195 (e) cos(3π + 3x) = −4 cos3 x + 3 cos x; −0.384 1 − cos2 x (f ) tan2 x = ; 0.172 cos2 x

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Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Chapter 1

(e) −b (i) 1/b

√ (d) ± 1 − a2

(b) b

√ (g) ±2b 1 − b2 (k) 1/c

(f ) −a (j) −1/a

(h) 2b2 − 1 (l) (1 − b)/2

33. (a) −a

us uf

34. (a) The distance is 36/360 = 1/10th of a great circle, so it is (1/10)2πr = 2, 513.27 mi. (b) 36/360 = 1/10 35. If the arc length is x, then solve the ratio

x 2πr = to get x ≈ 87, 458 km. 1 27.3

36. The distance travelled is equal to the length of that portion of the circumference of the wheel which touches the road, and that is the fraction 225/360 of a circumference, so a distance of (225/360)(2π)3 = 11.78 ft 37. The second quarter revolves twice (720◦ ) about its own center.

38. Add r to itself until you exceed 2πr; since 6r < 2πr < 7r, you can cut oﬀ 6 pieces of pie, but 3 2πr − 6r = 1 − times the original there’s not enough for a full seventh piece. Remaining pie is 2πr π pie. (c) y = −5 sin 4x

39. (a) y = 3 sin(x/2)

(b) y = 4 cos 2x

40. (a) y = 1 + cos πx

(b) y = 1 + 2 sin x

41. (a) y = sin(x + π/2)

(b) y = 3 + 3 sin(2x/9)

√ 42. V = 120 2 sin(120πt)

43. (a) 3, π/2, 0

(b) 2, 2, 0

y 3

1 c/2

2 x 2

1 x

-2

44. (a) 4, π, 0

-2

uh a

-4

-2

4c 6c

(b) 1/2, 2π/3, π/3

y 4

0.4

f l -0.2

x -2

-4

45. (a) A sin(ωt + θ) = A sin(ωt) cos θ + A cos(ωt) sin θ = A1 sin(ωt) + A2 cos(ωt) (b) A1 = A cos θ, A2 = A sin θ, so A = A21 + A22 and θ = tan−1 (A2 /A1 ).

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15c 21c 2 2

Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Exercise Set 1.7

√ 1 (c) A = 5 13/2, θ = tan−1 √ ; 2 3 √ 5 13 −1 1 √ sin 2πt + tan x= 2 2 3

46. three; x = 0, x = ±1.8955

us uf

–3

EXERCISE SET 1.7

appear appear appear appear

to to to to

be lie lie be

periodic, so a trigonometric model may be appropriate. near a parabola, so a quadratic model may be appropriate. near a line, so a linear model may be appropriate. randomly distributed, so no model seems to be appropriate.

data data data data

The The The The

1. The sum of the squares for the residuals for line I is approximately 12 + 12 + 12 + 02 + 22 + 12 + 12 + 12 = 10, and the same for line II is approximately 02 + (0.4)2 + (1.2)2 + 02 + (2.2)2 + (0.6)2 + (0.2)2 + 02 = 6.84; line II is the regression line. 2. (a) (b) (c) (d)

3. Least squares line S = 1.5388t − 2842.9, correlation coeﬃcient 0.83409 S

240 230 220 210 200 190

1980 1983 1986 1989 1992 1995 1998

4. (a) p = 0.0154T + 4.19

(b) 3.42 atm

5. (a) Least squares line p = 0.0146T + 3.98, correlation coeﬃcient 0.9999 (b) p = 3.25 atm (c) T = −272◦ C

uh a

6. (a) p = 0.0203 T + 5.54, r = 0.9999 (b) T = −273 (c) 1.05 p = 0.0203(T )(1.1) + 5.54 and p = 0.0203T + 5.54, subtract to get .05p = 0.1(0.0203T ), p = 2(0.0203T ). But p = 0.0203T + 5.54, equate the right hand sides, get T = 5.54/0.0203 ≈ 273◦ C (b) T = −214◦ C

7. (a) R = 0.00723T + 1.55

-10

-3

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Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Chapter 1

(b) T = −203◦ C

9. (a) S = 0.50179w − 0.00643

(b) S = 8, w = 16 lb

8. (a) y = 0.0236x + 4.79

us uf

10. (a) S = 0.756w − 0.0133 (b) 2S = 0.756(w + 5) − 0.0133, subtract to get S = 5(0.756) = 3.78

11. (a) Let h denote the height in inches and y the number of rebounds per minute. Then y = 0.00630h − 0.266, r = 0.313 y

(c)

0.3

No, the data points are too widely scattered.

(b)

0.25 0.2 0.15 0.1

0.05 h 66 67 68 69 70 71 72 73 74 75

(b)

12. (a) Let h denote the height in inches and y the weight in pounds. Then y = 7.73h − 391 y

280 260

240 220

180 h 72

160

200

13. (a) H ≈ 20000/110 ≈ 181 km/s/Mly (b) One light year is 9.408 × 1012 km and 1 d 1 9.408 × 1018 km t= = = = = 4.704 × 1017 s = 1.492 × 1010 years. v H 20km/s/Mly 20km/s (c) The Universe would be even older. (b) f = 85.7

15. (a) P = 0.322t2 + 0.0671t + 0.00837

(b) P = 1.43 cm

14. (a) f = 0.906m + 5.93

uh a

16. The population is modeled by P = 0.0045833t2 − 16.378t + 14635; in the year 2000 the population would be P = 212, 200, 000. This is far short; in 1990 the population of the US was approximately 250,000,000.

C 17. As in Example 4, a possible model is of the form T = D + A sin B t − . Since the longest B day has 993 minutes and the shortest has 706, take 2A = 993 − 706 = 287 or A = 143.5. The midpoint between the longest and shortest days is 849.5 minutes, so there is a vertical shift of D = 849.5. The period is about 365.25 days, so 2π/B = 365.25 or B = π/183. Note that the sine C function takes the value −1 when t − = −91.8125, and T is a minimum at about t = 0. Thus B π C π

is a model for the temperature. the phase shift ≈ 91.5. Hence T = 849.5 + 143.5 sin t− B 183 2

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Exercise Set 1.8

us uf

18. As in Example 4, a possible model for the fraction f of illumination is of the form

C . Since the greatest fraction of illumination is 1 and the least is 0, f = D + A sin B t − B 2A = 1, A = 1/2. The midpoint of the fraction of illumination is 1/2, so there is a vertical shift of D = 1/2. The period is approximately 30 days, so 2π/B = 30 or B = π/15. The phase shift is

π 49 approximately 49/2, so C/B = 49/2, and f = 1/2 + 1/2 sin t− 15 2 √ 19. t = 0.445 d

2.3

25 0

(b) 238,000 km

1. (a) x + 1 = t = y − 1, y = x + 2 y

2 t=0

x 2

2. (a) x2 + y 2 = 1 y t = 0.5

t = 0.25

t = 0.75

t=1 -1

3. t = (x + 4)/3; y = 2x + 10

0 1 t x −1 0 y 1 2

(c)

t 0 x 1 y 0

2 1 3

3 2 4

4 3 5

5 4 6

t=4 t=3 t=2 t=1

(c)

t=5

EXERCISE SET 1.8

20. (a) t = 0.373 r1.5

0.2500 0.50 0.7500 1 0.7071 0.00 −0.7071 −1 0.7071 1.00 0.7071 0

t=0 x 1

4. t = x + 3; y = 3x + 2, −3 ≤ x ≤ 0

uh a

-8

(0, 2) x

x 6

(-3, -7)

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Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Chapter 1

x -5

7. cos t = (x − 3)/2, sin t = (y − 2)/4; (x − 3)2 /4 + (y − 2)2 /16 = 1

6. t = x2 ; y = 2x2 + 4, x≥0

5. cos t = x/2, sin t = y/5; x2 /4 + y 2 /25 = 1

us uf

-5

-2

9. cos 2t = 1 − 2 sin2 t; x = 1 − 2y 2 , −1 ≤ y ≤ 1

8. sec2 t − tan2 t = 1; x2 − y 2 = 1, x ≤ −1 and y ≥ 0

10. t = (x − 3)/4; y = (x − 3)2 − 9 y

x -1

-1

12. y = x − 1, x ≥ 1, y ≥ 0 y

x 2

13. x = 5 cos t, y = −5 sin t, 0 ≤ t ≤ 2π

1 x 1

14. x = cos t, y = sin t, π ≤ t ≤ 3π/2 1

7.5

-1

-7.5

16. x = 2 cos t, y = 3 sin t, 0 ≤ t ≤ 2π

15. x = 2, y = t

uh a M

-1

–1

-5

(3, -9)

11. x/2 + y/3 = 1, 0 ≤ x ≤ 2, 0 ≤ y ≤ 3

-2

–3

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Exercise Set 1.8

17. x = t2 , y = t, −1 ≤ t ≤ 1

18. x = 1 + 4 cos t, y = −3 + 4 sin t, 0 ≤ t ≤ 2π

us uf

-7 1

-8

-1

IV, because x always increases whereas y oscillates. II, because (x/2)2 + (y/3)2 = 1, an ellipse. V, because x2 + y 2 = t2 increases in magnitude while x and y keep changing sign. VI; examine the cases t < −1 and t > −1 and you see the curve lies in the ﬁrst, second and fourth quadrants only. (e) III because y > 0. (f ) I; since x and y are bounded, the answer must be I or II; but as t runs, say, from 0 to π, x goes directly from 2 to −2, but y goes from 0 to 1 to 0 to −1 and back to 0, which describes I but not II.

19. (a) (b) (c) (d)

21. (a)

20. (a) from left to right (b) counterclockwise (c) counterclockwise (d) As t travels from −∞ to −1, the curve goes from (near) the origin in the third quadrant and travels up and left. As t travels from −1 to +∞ the curve comes from way down in the second quadrant, hits the origin at t = 0, and then makes the loop clockwise and finally approaches the origin again as t → +∞. (e) from left to right (f ) Starting, say, at (1, 0), the curve goes up into the first quadrant, loops back through the origin and into the third quadrant, and then continues the figure-eight.

-35

(b)

t 0 x 0 y 1

1 2 5.5 8 1.5 3

3 4 5 4.5 −8 −32.5 5.5 9 13.5

√ (d) for 0 < t < 2 2

√ (c) x = 0 when t = 0, 2 3. (e) at t = 2 22. (a)

(b) y is always ≥ 1 since cos t ≤ 1

uh a

-2

14 0

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Chapter 1

(b)

-1

-5

(b)

1.7

2.3

-10

-2.3

24. (a)

23. (a)

us uf

-1.7

x − x0 y − y0 = x1 − x0 y1 − y0 (b) Set t = 0 to get (x0 , y0 ); t = 1 for (x1 , y1 ).

25. (a) Eliminate t to get

(d) x = 2 − t, y = 4 − 6t

(b) x = at, y = b(1 − t), 0 ≤ t ≤ 1

26. (a) x = −3 − 2t, y = −4 + 5t, 0 ≤ t ≤ 1

(b) t = 1/2 28. x = 2 + t, y = −1 + 2t

27. (a) |R−P |2 = (x−x0 )2 +(y−y0 )2 = t2 [(x1 −x0 )2 +(y1 −y0 )2 ] and |Q−P |2 = (x1 −x0 )2 +(y1 −y0 )2 , so r = |R − P | = |Q − P |t = qt. (c) t = 3/4

(b) (9/4, −1/2)

(a) (5/2, 0)

29. The two branches corresponding to −1 ≤ t ≤ 0 and 0 ≤ t ≤ 1 coincide. t − t0 y − y0 y1 − y0 to obtain = . t1 − t0 x − x0 x1 − x0

30. (a) Eliminate

(b) from (x0 , y0 ) to (x1 , y1 ) 5

-2

uh a

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Exercise Set 1.8

31. (a)

y−d x−b = a c

(b)

y 3

1 x 2

32. (a) If a = 0 the line segment is vertical; if c = 0 it is horizontal. (b) The curve degenerates to the point (b, d). y 2

x = 1/2 − 4t, y = 1/2 x = −1/2, y = 1/2 − 4(t − 1/4) x = −1/2 + 4(t − 1/2), y = −1/2 x = 1/2, y = −1/2 + 4(t − 3/4)

(c)

35. (a) x = 4 cos t, y = 3 sin t (b) x = −1 + 4 cos t, y = 2 + 3 sin t

for 0 ≤ t ≤ 1/4 for 1/4 ≤ t ≤ 1/2 for 1/2 ≤ t ≤ 3/4 for 3/4 ≤ t ≤ 1

34.

x 0.5

us uf

33.

-4

-5

-3

3 -1

36. (a) t = x/(v0 cos α), so y = x tan α − gx2 /(2v02 cos2 α). y

(b)

10000

6000

x 40000 80000

uh a

2000

√ √ 37. (a) From Exercise 36, x = 400 2t, y = 400 2t − 4.9t2 . (b) 16,326.53 m (c) 65,306.12 m

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Chapter 1

(b)

-25

–15

(c)

–15 15

-25

38. (a)

us uf

-25

–15 a = 2, b = 3

–15 a = 3, b = 2

-25

–15 a = 2, b = 7

39. Assume that a = 0 and b = 0; eliminate the parameter to get (x − h)2 /a2 + (y − k)2 /b2 = 1. If |a| = |b| the curve is a circle with center (h, k) and radius |a|; if |a| = |b| the curve is an ellipse with center (h, k) and major axis parallel to the x-axis when |a| > |b|, or major axis parallel to the y-axis when |a| < |b|.

(a) ellipses with a ﬁxed center and varying axes of symmetry (b) (assume a = 0 and b = 0) ellipses with varying center and ﬁxed axes of symmetry (c) circles of radius 1 with centers on the line y = x − 1

40. Refer to the diagram to get bθ = aφ, θ = aφ/b but

θ − α = φ + π/2 so α = θ − φ − π/2 = (a/b − 1)φ − π/2 x = (a − b) cos φ − b sin α a−b = (a − b) cos φ + b cos φ, b

a−b

␪ ␾

uh a

y = (a − b) sin φ − b cos α a−b = (a − b) sin φ − b sin φ. b

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␣

b␪ a␾ x

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Chapter 1 Supplementary Exercises

41. (a)

us uf

-a

(b) Use b = a/4 in the equations of Exercise 40 to get x = 34 a cos φ + 14 a cos 3φ, y = 34 a sin φ − 14 a sin 3φ;

but trigonometric identities yield cos 3φ = 4 cos3 φ − 3 cos φ, sin 3φ = 3 sin φ − 4 sin3 φ, so x = a cos3 φ, y = a sin3 φ. (c) x2/3 + y 2/3 = a2/3 (cos2 φ + sin2 φ) = a2/3 y

y 5

-1

1 -3 a = 1/2

a=5

(x − a)2 + y 2 = (2a cos2 t − a)2 + (2a cos t sin t)2

(b)

x 5

-5

a=3 -1

42. (a)

= 4a2 cos4 t − 4a2 cos2 t + a2 + 4a2 cos2 t sin2 t

= 4a2 cos4 t − 4a2 cos2 t + a2 + 4a2 cos2 t(1 − cos2 t) = a2 ,

a circle about (a, 0) of radius a

CHAPTER 1 SUPPLEMENTARY EXERCISES 1. 1940-45; the greatest ﬁve-year slope

f (−1) = 3.3, g(3) = 2 x = −3, 3 x < −2, x > 3 the domain is −5 ≤ x ≤ 5 and the range is −5 ≤ y ≤ 4 the domain is −4 ≤ x ≤ 4.1, the range is −3 ≤ y ≤ 5 f (x) = 0 at x = −3, 5; g(x) = 0 at x = −3, 2

2. (a) (b) (c) (d) (e) (f )

uh a

t 4

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Chapter 1

5. If the side has length x and height h, then V = 8 = x2 h, so h = 8/x2 . Then the cost C = 5x2 + 2(4)(xh) = 5x2 + 64/x.

us uf

6. Assume that the paint is applied in a thin veneer of uniform thickness, so that the quantity of paint to be used is proportional to the area covered. If P is the amount of paint to be used, P = kπr2 . The constant k depends on physical factors, such as the thickness of the paint, absorption of the wood, etc. y

x -5

-1

8. Suppose the radius of the uncoated ball is r and that of the coated ball is r + h. Then the plastic has volume equal to the diﬀerence of the volumes, i.e. V = 43 π(r + h)3 − 43 πr3 = 43 πh[3r2 + 3rh + h2 ] in3 . 9. (a) The base has sides (10 − 2x)/2 and 6 − 2x, and the height is x, so V = (6 − 2x)(5 − x)x ft3 .

(b) From the picture we see that x < 5 and 2x < 6, so 0 < x < 3. (c) 3.57 ft ×3.79 ft ×1.21 ft

10. {x = 0} and ∅ (the empty set)

11. f (g(x)) = (3x + 2)2 + 1, g(f (x)) = 3(x2 + 1) + 2, so 9x2 + 12x + 5 = 3x2 + 5, 6x2 + 12x = 0, x = 0, −2 12. (a) (3 − x)/x

(b) no; f (g(x)) can be deﬁned at x = 1, whereas g, and therefore f ◦ g, requires x = 1 13. 1/(2 − x2 )

15.

14. g(x) = x2 + 2x

−4 −3 −2 −1 0 −1 2 1

x f (x) g(x)

4 −3 −2 −1 1 4 −4 −2 (g ◦ f )(x) −1 −3

0 1

uh a

−4

−2

−4 2

2 0

3 3 y

(b)

x -3

-1 -1

(b) a square is even (d) odd × odd = even

17. (a) even × odd = odd (c) even + odd is neither

−2 −3

y = |x − 1|, y = |(−x) − 1| = |x + 1|, y = 2|x + 1|, y = 2|x + 1| − 3, y = −2|x + 1| + 3

16. (a)

−3 −1 −4

(f ◦ g)(x)

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us uf

Ď&#x20AC; 3Ď&#x20AC; Ď&#x20AC; 5Ď&#x20AC; 7Ď&#x20AC; 11Ď&#x20AC; ,â&#x2C6;&#x2019; ,â&#x2C6;&#x2019; 18. (a) y = cos x â&#x2C6;&#x2019; 2 sin x cos x = (1 â&#x2C6;&#x2019; 2 sin x) cos x, so x = Âą , Âą , , 2 2 6 6 6 6 Ď&#x20AC; â&#x2C6;&#x161; Ď&#x20AC; 5Ď&#x20AC; â&#x2C6;&#x161; 3Ď&#x20AC; 7Ď&#x20AC; â&#x2C6;&#x161; 11Ď&#x20AC; â&#x2C6;&#x161; , 3/2 , , â&#x2C6;&#x2019; 3/2 , â&#x2C6;&#x2019; , â&#x2C6;&#x2019; 3/2 , â&#x2C6;&#x2019; , 3/2 (b) Âą , 0 , Âą , 0 , 2 2 6 6 6 6

Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Chapter 1 Supplementary Exercises

19. (a) If x denotes the distance from A to the base of the tower, and y the distance from B to the base, then x2 +d2 = y 2 . Moreover h = x tan Îą = y tan Î˛, so d2 = y 2 â&#x2C6;&#x2019;x2 = h2 (cot2 Î˛â&#x2C6;&#x2019;cot2 Îą), d2 d2 tan2 Îą tan2 Î˛ d2 = = , which yields the result. h2 = 2 2 2 2 cot Î˛ â&#x2C6;&#x2019; cot Îą 1/ tan Î˛ â&#x2C6;&#x2019; 1/ tan Îą tan2 Îą â&#x2C6;&#x2019; tan2 Î˛

(b) 295.72 ft. y

20. (a) 60

20 t 100

300

-20

3Ď&#x20AC; 2Ď&#x20AC; (t â&#x2C6;&#x2019; 101) = , or t = 374.75, which is the same date as t = 9.75, so during the 365 2 night of January 10th-11th

(b) when

21. When x = 0 the value of the green curve is higher than that of the blue curve, therefore the blue curve is given by y = 1 + 2 sin x. The points A, B, C, D are the points of intersection of the two curves, i.e. where 1+2 sin x = 2 sin(x/2)+2 cos(x/2). Let sin(x/2) = p, cos(x/2) = q. Then 2 sin x = 4 sin(x/2) cos(x/2), so the equation which yields the points of intersection becomes 1 + 4pq = 2p + 2q, 4pq â&#x2C6;&#x2019; 2p â&#x2C6;&#x2019; 2q + 1 = 0, (2p â&#x2C6;&#x2019; 1)(2q â&#x2C6;&#x2019; 1) = 0; thus whenever either sin(x/2) =â&#x2C6;&#x161;1/2 or cos(x/2) = 1/2, i.e. whenâ&#x2C6;&#x161;x/2 = Ď&#x20AC;/6, 5Ď&#x20AC;/6, ÂąĎ&#x20AC;/3. Thus A â&#x2C6;&#x161; has coordinates (â&#x2C6;&#x2019;2Ď&#x20AC;/3, 1 â&#x2C6;&#x2019; 3), B has â&#x2C6;&#x161; coordinates (Ď&#x20AC;/3, 1 + 3), C has coordinates (2Ď&#x20AC;/3, 1 + 3), and D has coordinates (5Ď&#x20AC;/3, 1 â&#x2C6;&#x2019; 3).

22. Let y = A + B sin(at + b). Since the maximum and minimum values of y are 35 and 5, A + B = 35 and A â&#x2C6;&#x2019; B = 5, A = 20, B = 15. The period is 12 hours, so 12a = 2Ď&#x20AC; and a = Ď&#x20AC;/6. The maximum occurs at t = 2, so 1 = sin(2a + b) = sin(Ď&#x20AC;/3 + b), Ď&#x20AC;/3 + b = Ď&#x20AC;/2, b = Ď&#x20AC;/2 â&#x2C6;&#x2019; Ď&#x20AC;/3 = Ď&#x20AC;/6 and y = 20 + 15 sin(Ď&#x20AC;t/6 + Ď&#x20AC;/6).

23. (a) The circle of radius 1 centered at (a, a2 ); therefore, the family of all circles of radius 1 with centers on the parabola y = x2 . (b) All parabolas which open up, have latus rectum equal to 1 and vertex on the line y = x/2. y

25. 1

x -2

-2

uh a

24. (a) x = f (1 â&#x2C6;&#x2019; t), y = g(1 â&#x2C6;&#x2019; t)

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Chapter 1

26. Let y = ax2 + bx + c. Then 4a + 2b + c = 0, 64a + 8b + c = 18, 64a − 8b + c = 18, from which b = 0 3 2 x − 65 . and 60a = 18, or ﬁnally y = 10

27.

us uf

2 1 x -1

-1

-2

√ (x − 1)2 + ( x − 2)2 ;

30. d =

d = 9.1 at x = 1.358094

(x − 1)2 + 1/x2 ;

d = 0.82 at x = 1.380278

29. d =

28. (a) R = R0 is the R-intercept, R0 k is the slope, and T = −1/k is the T -intercept (b) −1/k = −273, or k = 1/273 (c) 1.1 = R0 (1 + 20/273), or R0 = 1.025 (d) T = 126.55◦ C

y 2

1.6

1.2 1 0.8 0.5 1

x 2

32. (a)

31. w = 63.9V , w = 63.9πh2 (5/2 − h/3); h = 0.48 ft when w = 108 lb (b)

W 3000

w = 63.9πh2 (5/2 − h/3); at h = 5/2, w = 2091.12 lb

1000 3

33. (a)

34. (a)

200

uh a

100

10 t 30

(b) T = 17◦ F, 27◦ F, 32◦ F

(b) N = 80 when t = 9.35 yrs (c) 220 sheep

v 20

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Chapter 1 Supplementary Exercises

(b) T = 3◦ F, −11◦ F, −18◦ F, −22◦ F

WCI 20

35. (a)

us uf

v 20

-20

36. The domain is the set of all x, the range is −0.1746 ≤ y ≤ 0.1227.

37. The domain is the set −0.7245 ≤ x ≤ 1.2207, the range is −1.0551 ≤ y ≤ 1.4902. 38. (a) The potato is done in the interval 27.65 < t < 32.71. (b) 91.54 min.

39. (a)

5 t 1

(b) As t → ∞, (0.273)t → 0, and thus v → 24.61 ft/s. (c) For large t the velocity approaches c.

(d) No; but it comes very close (arbitrarily close). (e) 3.013 s

41. (a)

40. (a) y = −0.01716428571x + 1.433827619 1.90

1.92

1.94

3.4161

3.4639

3.5100

(b) y = 1.9589x − 0.2910

1.96

1.98

2.00

2.02

2.04

2.06

2.08

2.10

3.5543

3.5967

3.6372

3.6756

3.7119

3.7459

3.7775

3.8068

(c) y − 3.6372 = 1.9589(x − 2), or y = 1.9589x − 0.2806 (d) As one zooms in on the point (2, f (2)) the two curves seem to converge to one line.

uh a

3.8

42. (a)

−0.10 0.9950

1.9

2.2 3.4

−0.08 0.9968

−0.06 0.9982

−0.04 0.9992

−0.02 0.9998

0.00 1.0000

0.02 0.9998

0.04 0.9992

0.06 0.9982

0.08 0.9968

0.10 0.9950

(b) y = − 12 x2 + 1

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Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Chapter 1

us uf

1.2

1.7

-0.8

43. The data are periodic, so it is reasonable that a trigonometric function might approximate them.

C . Since the highest level is 1.032 A possible model is of the form T = D + A sin B t − B meters and the lowest is 0.045, take 2A = 1.032 − 0.042 = 0.990 or A = 0.495. The midpoint between the lowest and highest levels is 0.537 meters, so there is a vertical shift of D = 0.537. C The period is about 12 hours, so 2π/B = 12 or B = π/6. The phase shift ≈ 6.5. Hence B

π (t − 6.5) is a model for the temperature. T = 0.537 + 0.495 sin 6

T 1. 0.8

0.4 0.2 t

10 20 30 40 50 60 70

0.6

CHAPTER 1 HORIZON MODULE

1. (a) 0.25, 6.25 × 10−2 , 3.91 × 10−3 , 1.53 × 10−5 , 2.32 × 10−10 , 5.42 × 10−20 , 2.94 × 10−39 , 8.64 × 10−78 , 7.46 × 10−155 , 5.56 × 10−309 ; 1, 1, 1, 1, 1, 1, 1, 1, 1, 1; 4, 16, 256, 65536, 4.29 × 109 , 1.84 × 1019 , 3.40 × 1038 , 1.16 × 1077 , 1.34 × 10154 , 1.80 × 10308

(b) yn =

1 2n

yn+1 = 1.05yn y0 =$1000, y1 =$1050, y2 =$1102.50, y3 =$1157.62, y4 =$1215.51, y5 =$1276.28 yn+1 = 1.05yn for n ≥ 1 yn = (1.05)n 1000; y15 =$2078.93

uh a

4. (a) (b) (c) (d)

1 1 1 1 1 1 , , , , , 2 4 8 16 32 64

3. (a)

2. 1, 3, 2.3333333, 2.23809524, 2.23606890, 2.23606798, . . .

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Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Chapter 1 Horizon Module

5. (a) x1/2 , x1/4 , x1/8 , x1/16 , x1/32 (b) They tend to the horizontal line y = 1, with a hole at x = 0.

us uf

1.8

(c)

2 3

3 5

5 8

8 13

13 21

21 34

34 55

55 89

89 144

1, 2, 3, 5, 8, 13, 21, 34, 55, 89; each new numerator is the sum of the previous two numerators.

(b)

1 2

144 233 377 610 987 1597 2584 4181 6765 10946 , , , , , , , , , 233 377 610 987 1597 2584 4181 6765 10946 17711

(d) F0 = 1, F1 = 1, Fn = Fn−1 + Fn−2 for n ≥ 2. (e) the positive solution 7. (a) y1 = cr, y2 = cy1 = cr2 , y3 = cr3 , y4 = cr4

6. (a)

(b) yn = crn

8. The ﬁrst point on the curve is (c, kc(1 − c)), so y1 = kc(1 − c) and hence y1 is the ﬁrst iterate. The point on the line to the right of this point has equal coordinates (y1 , y1 ), and so the point above it on the curve has coordinates (y1 , ky1 (1 − y1 )); thus y2 = ky1 (1 − y1 ), and y2 is the second iterate, etc. 9. (a) 0.261, 0.559, 0.715, 0.591, 0.701

uh a

(b) It appears to approach a point somewhere near 0.65.

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Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH CHAPTER 2

Limits and Continuity

us uf

EXERCISE SET 2.1 (b) 3 (e) −1

2. (a) 2 (d) 2

(b) 0 (e) 0

3. (a) 1

(b) 1

(d) 1

(e) −∞

(f ) +∞

4. (a) 3

(b) 3

(d) 3

(e) +∞

(f ) +∞

5. (a) 0

(b) 0

(d) 3

1. (a) −1 (d) 1

(f ) +∞

6. (a) 2

(b) 2

(d) 3

(e) −∞

(f ) +∞

(b) +∞

9. (a) −∞

(b) −∞

12. (a) 3 (d) 3

(d) 1

(e) 1

(f ) 2

(b) 0 (e) does not exist

(b) 3 (e) does not exist

14. for all x0 = −6, 3

2 1.5 1.1 1.01 0.1429 0.2105 0.3021 0.3300

uh a

(f ) −1

(b) −∞ (e) +∞

13. for all x0 = −4

11. (a) 0 (d) 0

19. (a)

(e) 0

10. (a) 1 (d) −2

(d) undef

8. (a) +∞

(b) +∞ (e) 2

7. (a) −∞ (d) undef

(e) +∞

1.001 0 0.5 0.9 0.3330 1.0000 0.5714 0.3690

0.99 0.999 0.3367 0.3337

The limit is 1/3.

2 0

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Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Exercise Set 2.1

2 1.5 1.1 1.01 1.001 1.0001 0.4286 1.0526 6.344 66.33 666.3 6666.3 The limit is +∞.

us uf

(b)

(c)

0.9999 0 0.5 0.9 0.99 0.999 −1 −1.7143 −7.0111 −67.001 −667.0 −6667.0 0

The limit is −∞.

-50

−0.25 −0.1 −0.001 −0.0001 0.0001 0.001 0.5359 0.5132 0.5001 0.5000 0.5000 0.4999

20. (a)

0.6

0.1 0.25 0.4881 0.4721

The limit is 1/2.

-0.25

0.25

0.25 0.1 0.001 0.0001 8.4721 20.488 2000.5 20001 The limit is +∞.

100

(b)

0.25

uh a

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Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Chapter 2

−0.25 −0.1 −0.001 −0.0001 −7.4641 −19.487 −1999.5 −20000 0 -0.25

(c)

The limit is −∞.

us uf

The limit is 3.

-0.25

0.25 2

0 1

−0.5 −0.9 −0.99 −0.999 −1.5 −1.1 −1.01 −1.001 1.7552 6.2161 54.87 541.1 −0.1415 −4.536 −53.19 −539.5

(b)

0.25 0.001 0.1 3.0000 2.9552 2.7266

−0.25 −0.1 −0.001 −0.0001 0.0001 3.0000 3.0000 2.7266 2.9552 3.0000

21. (a)

-100

The limit does not exist.

-1.5

-60

0 −0.5 −0.9 −0.99 1.5574 1.0926 1.0033 1.0000 1.5

−1.001 1.0000

0 1

uh a

-1.5

−0.999 −1.5 −1.1 −1.01 1.0000 1.0926 1.0033 1.0000 The limit is 1.

22. (a)

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Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Exercise Set 2.1

−0.25 −0.1 −0.001 −0.0001 0.0001 0.001 1.9794 2.4132 2.5000 2.5000 2.5000 2.5000

2.5

0.1 0.25 2.4132 1.9794

The limit is 5/2.

-0.25

us uf

(b)

0.25

23. The height of the ball at time t = 0.25 + ∆t is s(0.25 + ∆t) = −16(0.25 + ∆t)2 + 29(0.25 + ∆t) + 6, so the distance traveled over the interval from t = 0.25 − ∆t to t = 0.25 + ∆t is s(0.25 + ∆t) − s(0.25 − ∆t) = −64(0.25)∆t + 58∆t. Thus the average velocity over the same interval is given by vave = [s(0.25 + ∆t) − s(0.25 − ∆t)]/2∆t = (−64(0.25)∆t + 58∆t)/2∆t = 21 ft/s, and this will also be the instantaneous velocity, since it happens to be independent of ∆t.

−10 −100,000,000 −100,000 −1000 −100 2.0000 2.0001 2.0050 2.0521 2.8333

100,000 100,000,000 2.0000 2.0000

10 100 1000 1.6429 1.9519 1.9950

25. (a)

24. The height of the ball at time t = 0.75 + ∆t is s(0.75 + ∆t) = −16(0.75 + ∆t)2 + 29(0.75 + ∆t) + 6, so the distance traveled over the interval from t = 0.75 − ∆t to t = 0.75 + ∆t is s(0.75 + ∆t) − s(0.75 − ∆t) = −64(0.75)∆t + 58∆t. Thus the average velocity over the same interval is given by vave = [s(0.75 + ∆t) − s(0.75 − ∆t)]/2∆t = (−64(0.75)∆t + 58∆t)/2∆t = 5 ft/s, and this will also be the instantaneous velocity, since it happens to be independent of ∆t.

-14

asymptote y = 2 as x → ±∞

-40

−100,000,000 −100,000 −1000 20.0855 20.0864 20.1763

(b)

−100 −10 10 21.0294 35.4013 13.7858

100 19.2186

1000 19.9955

asymptote y = 20.086.

uh a

100,000 100,000,000 20.0846 20.0855

-160

160 0

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Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Chapter 2

−100,000

−1000

−100

−10

100

1000

100,000

100,000,000

−100,000,001

−100,000

−1001

−101.0

−11.2

9.2

99.0

999.0

99,999

99,999,999

−100,000,000

no horizontal asymptote

-20

us uf

(c)

–50

−100,000,000

−100,000

−1000

−100

−10

100

1000

100,000

100,000,000

0.2000

0.1976

0.2000

0.2

asymptote y = 1/5 as x → ±∞ 10

-10

26. (a)

-1.2

1000 1.77 × 10301

100,000 ?

100,000,000 ?

−10 −100,000,000 −100,000 −1000 −100 0.0000 0.0000 0.0016 0.0000 0.0000

10 100 1668.0 2.09 × 1018

(b)

asymptote y = 0 as x → −∞, none as x → +∞

-6

−100,000,000 −100,000 −1000 −100 −10 10 100 0.0000 0.0000 0.0008 −0.0051 −0.0544 −0.0544 −0.0051

(c)

uh a

1000 100,000 0.0008 0.0000 1.1

-30

100,000,000 0.0000 asymptote y = 0 as x → ±∞

30 -0.3

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Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Exercise Set 2.2

27. It appears that lim n(t) = +∞, and lim e(t) = c. t→+∞

29. (a)

lim

t→0+

sin t t

(b)

31.

t−1 t+1

(c)

1 (b) lim+ t→0 t + 1

(c)

lim f (x) = L and lim = L

x→−∞

lim (1 + 2t)1/t

t→0−

lim

t→0−

2 1+ t

cos πt 30. (a) lim + πt t→0

lim

t→0+

28. (a) It is the initial temperature of the oven. (b) It is the ambient temperature, i.e. the temperature of the room.

us uf

t→+∞

x→+∞

32. (a) no (b) yes; tan x and sec x at x = nπ + π/2, and cot x and csc x at x = nπ, n = 0, ±1, ±2, . . . 33. (a) The limit appears to be 3.

(b) The limit appears to be 3.

3.5

1 2.5

0.001 2.5

3.5

– 0.001

–1

– 0.000001

0.000001

2.5

35. (a) The plot over the interval [−a, a] becomes subject to catastrophic subtraction if a is small enough (the size depending on the machine). (c) It does not.

EXERCISE SET 2.2

(b) π

(d) 36

2. (a) 1

(b) −1

(d) −1

uh a

1. (a) 7

3. (a) −6 (b) 13 (c) −8 (d) 16 (e) 2 (f ) −1/2 (g) The limit doesn’t exist because the denominator tends to zero but the numerator doesn’t. (h) The limit doesn’t exist because the denominator tends to zero but the numerator doesn’t.

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Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Chapter 2

0 The limit doesn’t exist because lim f doesn’t exist and lim g does. 0 (d) 3 (e) 0 The limit doesn’t exist because the denominator tends to zero but the numerator doesn’t. The limit doesn’t exist because f (x) is not deﬁned for 0 ≤ x < 2. 1

4. (a) (b) (c) (f ) (g) (h) 5. 0

6. 3/4

8. −3

7. 8

us uf

10. 12

11. −4/5

12. 0

13. 3/2

14. 4/3

15. +∞

16. −∞

17. does not exist

18. +∞

19. −∞

20. does not exist

21. +∞

22. −∞

23. does not exist

24. −∞

25. +∞

26. does not exist

27. +∞

28. +∞

(b) 2

34. (a) −2

(b) 0

33. (a) 2

4 x 1

(b) F (x) = x − 3

36. (a) −6

(b)

35. (a) 3

32. −19

30. 4

29. 6

9. 4

lim

x→0−

= lim

x→0−

x+1 = +∞ x2

39.

lim

x→0

√

x 1 = 4 x+4+2

x2 =0 x+4+2

uh a

38.

1 1 + x x2

37. (a) Theorem 2.2.2(a) doesn’t apply; moreover one cannot add/subtract inﬁnities. 1 x−1 1 (b) lim = lim = −∞ − x x2 x2 x→0+ x→0+

40.

lim

x→0

√

41. The left and/or right limits could be plus or minus inﬁnity; or the limit could exist, or equal any preassigned real number. For example, let q(x) = x − x0 and let p(x) = a(x − x0 )n where n takes on the values 0, 1, 2.

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Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Exercise Set 2.3

EXERCISE SET 2.3 (b) −1

2. (a) 1

(b) −∞

us uf

1. (a) −3

3. (a) −12 (b) 21 (c) −15 (d) 25 (e) 2 (f ) −3/5 (g) 0 (h) The limit doesn’t exist because the denominator tends to zero but the numerator doesn’t.

5. +∞

7. −∞

6. 5 10. +∞

11. 3/2

13. 0

14. 0

15. 0

17. −51/3 /2

18.

√ 21. 1/ 6

√ 22. −1/ 6

23.

25. −∞

26. +∞

27. −1/7

(b) −5

29. (a) +∞

√ 19. − 5

3/2

√

12. 5/2

8. +∞

16. 5/3

9. +∞

(d) −∞ (h) −7/12

(b) 0 (f ) −6/7

4. (a) 20 (e) −421/3

30. (a) 0

20.

24.

√ √

5 3

28. 4/7 (b) −6

√ x2 + 3 + x 3 2 31. lim ( x + 3 − x) √ = lim √ =0 2 2 x→+∞ x→+∞ x +3+x x +3+x √ x2 − 3x + x −3x 2 32. lim ( x − 3x − x) √ = lim √ = −3/2 2 2 x→+∞ x→+∞ x − 3x + x x − 3x + x

x→+∞

lim

x→+∞

√x2 + ax + x ax x2 + ax − x √ = lim √ = a/2 x2 + ax + x x→+∞ x2 + ax + x x2

+ ax −

√x2 + ax + √x2 + bx (a − b)x a−b √ √ + bx √ = lim √ = 2 x2 + ax + x2 + bx x→+∞ x2 + ax + x2 + bx

lim p(x) = (−1)n ∞ and lim p(x) = +∞

x→+∞

x→−∞

35.

34.

lim

33.

uh a

36. If m > n the limits are both zero. If m = n the limits are both 1. If n > m the limits are (−1)n+m ∞ and +∞, respectively. 37. If m > n the limits are both zero. If m = n the limits are both equal to am , the leading coeﬃcient of p. If n > m the limits are ±∞ where the sign depends on the sign of am and whether n is even or odd.

38. (a) p(x) = q(x) = x (c) p(x) = x2 , q(x) = x

(b) p(x) = x, q(x) = x2 (d) p(x) = x + 3, q(x) = x

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Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Chapter 2

39. If m > n the limit is 0. If m = n the limit is −3. If m < n and n − m is odd, then the limit is +∞; if m < n and n − m is even, then the limit is −∞.

41. f (x) = x + 2 +

us uf

40. If m > n the limit is zero. If m = n the limit is cm /dm . If n > m the limit is ±∞, where the sign depends on the signs of cn and dm . 2 , so lim (f (x) − (x + 2)) = 0 and f (x) is asymptotic to y = x + 2. x→±∞ x−2

-2

-20

42. f (x) = x2 − 1 + 3/x, so lim [f (x) − (x2 − 1) = 0] and f (x) is asymptotic to y = x2 − 1. x→±∞

-3

43. f (x) = −x2 + 1 + 2/(x − 3) so lim [f (x) − (−x2 + 1)] = 0 and f (x) is asymptotic to y = −x2 + 1. x→±∞

-30

3 3 − so lim [f (x) − x3 ] = 0 and f (x) is asymptotic to y = x3 . x→±∞ 2(x − 1) 2(x + 1) 30

-30

uh a

-1

44. f (x) = x3 +

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Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Exercise Set 2.4

45. f (x) − sin x = 0 and f (x) is asymptotic to y = sin x.

-3

-5

us uf

46. Note that the function is not deﬁned for −1 < x <= 1. For x outside this interval we have 2 f (x) = x2 + which suggests that lim [f (x) − |x| ] = 0 (this can be checked with a CAS) x→±∞ x−1 and hence f (x) is asymptotic to y = |x| .

-3

3 -1

EXERCISE SET 2.4

1. (a) |f (x) − f (0)| = |x + 2 − 2| = |x| < 0.1 if and only if |x| < 0.1 (b) |f (x) − f (3)| = |(4x − 5) − 7| = 4|x − 3| < 0.1 if and only if |x − 3| < (0.1)/4 = 0.0025 (c) |f (x) − f (4)| = |x2 − 16| < if |x − 4| < δ. We get f (x) = 16 + = 16.001 at x = 4.000124998, which corresponds to δ = 0.000124998; and f (x) = 16 − = 15.999 at x = 3.999874998, for which δ = 0.000125002. Use the smaller δ: thus |f (x) − 16| < provided |x − 4| < 0.000125 (to six decimals).

2. (a) |f (x) − f (0)| = |2x + 3 − 3| = 2|x| < 0.1 if and only if |x| < 0.05 (b) |f (x) − f (0)| = |2x + 3 − 3| = 2|x| < 0.01 if and only if |x| < 0.005 (c) |f (x) − f (0)| = |2x + 3 − 3| = 2|x| < 0.0012 if and only if |x| < 0.0006 3. (a) x1 = (1.95)2 = 3.8025, x2 = (2.05)2 = 4.2025 (b) δ = min ( |4 − 3.8025|, |4 − 4.2025| ) = 0.1975

4. (a) x1 = 1/(1.1) = 0.909090 . . . , x2 = 1/(0.9) = 1.111111 . . . (b) δ = min( |1 − 0.909090|, |1 − 1.111111| ) = 0.0909090 . . .

5. |(x3 −4x+5)−2| < 0.05, −0.05 < (x3 −4x+5)−2 < 0.05, 1.95 < x3 −4x+5 < 2.05; x3 −4x+5 = 1.95 at x = 1.0616, x3 − 4x + 5 = 2.05 at x = 0.9558; δ = min (1.0616 − 1, 1 − 0.9558) = 0.0442

uh a

2.2

0.9 1.9

1.1

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Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Chapter 2

√

5x + 1 = 3.5 at x = 2.25,

√

5x + 1 = 4.5 at x = 3.85, so δ = min(3 − 2.25, 3.85 − 3) = 0.75

us uf

7. With the TRACE feature of a calculator we discover that (to ﬁve decimal places) (0.87000, 1.80274) and (1.13000, 2.19301) belong to the graph. Set x0 = 0.87 and x1 = 1.13. Since the graph of f (x) rises from left to right, we see that if x0 < x < x1 then 1.80274 < f (x) < 2.19301, and therefore 1.8 < f (x) < 2.2. So we can take δ = 0.13. 8. From a calculator plot we conjecture that lim f (x) = 2. Using the TRACE feature we see that

x→1

the points (±0.2, 1.94709) belong to the graph. Thus if −0.2 < x < 0.2 then 1.95 < f (x) ≤ 2 and hence |f (x) − L| < 0.05 < 0.1 = .

9. |2x − 8| = 2|x − 4| < 0.1 if |x − 4| < 0.05, δ = 0.05

10. |x/2 + 1| = (1/2)|x − (−2)| < 0.1 if |x + 2| < 0.2, δ = 0.2

12. |5x − 2 − 13| = 5|x − 3| < 0.01 if |x − 3| <

1 700 ,

δ=

1 700

11. |7x + 5 − (−2)| = 7|x − (−1)| < 0.01 if |x + 1| < 1 500 ,

δ=

1 500

2 2 x − 4 − 4x + 8 x − 4 = |x − 2| < 0.05 if |x − 2| < 0.05, δ = 0.05 − 4 = 13. x−2 x−2

2 2 x − 1 x − 1 + 2x + 2 = |x + 1| < 0.05 if |x + 1| < 0.05, δ = 0.05 14. − (−2) = x+1 x+1 15. if δ < 1 then x2 − 16 = |x − 4||x + 4| < 9|x − 4| < 0.001 if |x − 4| <

1 9000 ,

δ=

1 9000

√ x + 3 √ |x − 9| |x − 9| 1 = √ < √ 16. if δ < 1 then | x − 3| √ < |x − 9| < 0.001 if |x − 9| < 0.004, 4 x+3 | x + 3| 8+3 δ = 0.004 1 1 |x − 5| |x − 5| ≤ < 0.05 if |x − 5| < 1, δ = 1 17. if δ ≤ 1 then − = x 5 5|x| 20

18. |x − 0| = |x| < 0.05 if |x| < 0.05, δ = 0.05 19. |3x − 15| = 3|x − 5| < if |x − 5| < 13 , δ = 13

uh a

20. |(4x − 5) − 7| = |4x − 12| = 4|x − 3| < if |x − 3| < 14 , δ = 14 21. |2x − 7 − (−3)| = 2|x − 2| < if |x − 2| < 12 , δ = 12 22. |2 − 3x − 5| = 3|x + 1| < if |x + 1| < 13 , δ = 13

2 x + x − 1 = |x| < if |x| < , δ = 23. x

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Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Exercise Set 2.4

us uf

25. if δ < 1 then |2x2 − 2| = 2|x − 1||x + 1| < 6|x − 1| < if |x − 1| < 16 , δ = min(1, 16 )

2 x − 9 24. − (−6) = |x + 3| < if |x + 3| < , δ = x+3

26. if δ < 1 then |x2 − 5 − 4| = |x − 3||x + 3| < 7|x − 3| < if |x − 3| < 17 , δ = min(1, 17 ) 27. if δ <

3|x − 13 | 1 1 < 18 x − then − 3 = x |x| 6

1 x − < if 3

1 < 3

1 18 ,

δ = min

1 1 , 6 18

28. If δ < 12 and |x − (−2)| < δ then − 52 < x < − 32 , x + 1 < − 12 , |x + 1| > 12 ; then 1 |x + 2| 1 1 1 x + 1 − (−1) = |x + 1| < 2|x + 2| < if |x + 2| < 2 , δ = min 2 , 2

√ √ √ x + 2 x − 4 1 √ 29. | x − 2| = ( x − 2) √ = < |x − 4| < if |x − 4| < 2 , δ = 2 x + 2 x + 2 2 √ x + 3 + 3 √ = √ |x − 6| ≤ 1 |x − 6| < if |x − 6| < 3 , δ = 3 30. | x + 3 − 3| √ 3 x + 3 + 3 x+3+3

31. |f (x) − 3| = |x + 2 − 3| = |x − 1| < if 0 < |x − 1| < , δ =

32. If δ < 1 then |(x2 + 3x − 1) − 9| = |(x − 2)(x + 5)| < 8|x − 2| < if |x − 2| < 18 , δ = min (1, 18 ) √ √ 1 < 0.1 if x > 10, N = 10 x2 1 x < 0.01 if x + 1 > 100, N = 99 − 1 = (b) |f (x) − L| = x + 1 x+1 1 1 (c) |f (x) − L| = 3 < if |x| > 10, x < −10, N = −10 x 1000 x 1 < 0.01 if |x + 1| > 100, −x − 1 > 100, x < −101, (d) |f (x) − L| = − 1 = x+1 x + 1 N = −101

35. (a)

(b)

(c)

1 1 < 0.1, x > 101/3 , N = 101/3 (b) 3 < 0.01, x > 1001/3 , N = 1001/3 x3 x 1 < 0.001, x > 10, N = 10 x3 x21 x22 1− 1− ; = 1 − , x = − = 1 − , x = 1 2 1 + x21 1 + x22 1− 1− (c) N = − N=

34. (a)

33. (a) |f (x) − L| =

(b) N = 1/ 3

uh a

36. (a) x1 = −1/ 3 ; x2 = 1/ 3 37.

1 < 0.01 if |x| > 10, N = 10 x2

1 < 0.005 if |x + 2| > 200, x > 198, N = 198 x+2 x 1 < 0.001 if |x + 1| > 1000, x > 999, N = 999 39. − 1 = x+1 x + 1

38.

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Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Chapter 2

1 42. 2 < 0.01 if |x| > 10, −x > 10, x < −10, N = −10 x

us uf

1 41. − 0 < 0.005 if |x + 2| > 200, −x − 2 > 200, x < −202, N = −202 x+2

11 4x − 1 < 0.1 if |2x + 5| > 110, 2x > 105, N = 52.5 − 2 = 40. 2x + 5 2x + 5

4x − 1 11 < 0.1 if |2x + 5| > 110, −2x − 5 > 110, 2x < −115, x < −57.5, N = −57.5 43. − 2 = 2x + 5 2x + 5 x 1 < 0.001 if |x + 1| > 1000, −x − 1 > 1000, x < −1001, N = −1001 44. − 1 = x+1 x + 1 1 1 1 45. 2 < if |x| > √ , N = √ x

1 1 1 1 1 46. < if |x| > , −x > , x < − , N = − x

1 < if |x + 2| > 1 , −x − 2 < 1 , x > −2 − 1 , N = −2 − 1 47. x + 2 1 < if |x + 2| > 1 , x + 2 > 1 , x > 1 − 2, N = 1 − 2 48. x + 2

x 1 < if |x + 1| > 1 , x > 1 − 1, N = 1 − 1 49. − 1 = x+1 x + 1

x 1 < if |x + 1| > 1 , −x − 1 > 1 , x < −1 − 1 , N = −1 − 1 50. − 1 = x+1 x + 1

11 4x − 1 < if |2x + 5| > 11 , −2x − 5 > 11 , 2x < − 11 − 5, x < − 11 − 5 , − 2 = 51. 2x + 5 2x + 5 2 2

(c)

1 1 > 100 if |x| < 2 x 10 −1 1 < −1000 if |x − 3| < √ (x − 3)2 10 10

1 1 > 1000 if |x − 1| < |x − 1| 1000 1 1 1 (d) − 4 < −10000 if x4 < , |x| < x 10000 10 (b)

53. (a)

5 11 N =− − 2 2 4x − 1 11 < if |2x + 5| > 11 , 2x > 11 − 5, x > 11 − 5 , N = 11 − 5 52. − 2 = 2x + 5 2x + 5 2 2 2 2

1 1 > 10 if and only if |x − 1| < √ (x − 1)2 10 1 1 > 1000 if and only if |x − 1| < √ (b) (x − 1)2 10 10 1 1 √ > 100000 if and only if |x − 1| < (c) 2 (x − 1) 100 10

uh a

54. (a)

55. if M > 0 then

1 1 1 1 , 0 < |x − 3| < √ , δ = √ > M , 0 < (x − 3)2 < 2 (x − 3) M M M

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Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Exercise Set 2.4

1 1 −1 1 ,δ=√ < M , 0 < (x − 3)2 < − , 0 < |x − 3| < √ (x − 3)2 M −M −M

57. if M > 0 then

1 1 1 > M , 0 < |x| < ,δ= |x| M M

58. if M > 0 then

1 1 1 > M , 0 < |x − 1| < ,δ= |x − 1| M M

60. if M > 0 then

1 1 1 1 > M , 0 < x4 < , x < 1/4 , δ = 1/4 x4 M M M

61. if x > 2 then |x + 1 − 3| = |x − 2| = x − 2 < if 2 < x < 2 + , δ =

us uf

1 1 1 1 < M , 0 < x4 < − , |x| < ,δ= M x4 (−M )1/4 (−M )1/4

59. if M < 0 then −

56. if M < 0 then

64. if x < 0 then

√ √

x − 4 < if x − 4 < 2 , 4 < x < 4 + 2 , δ = 2 −x < if −x < 2 , − 2 < x < 0, δ = 2

63. if x > 4 then

62. if x < 1 then |3x + 2 − 5| = |3x − 3| = 3|x − 1| = 3(1 − x) < if 1 − x < 13 , 1 − 13 < x < 1, δ = 13

65. if x > 2 then |f (x) − 2| = |x − 2| = x − 2 < if 2 < x < 2 + , δ =

66. if x < 2 then |f (x) − 6| = |3x − 6| = 3|x − 2| = 3(2 − x) < if 2 − x < 13 , 2 − 13 < x < 2, δ = 13

1 1 1 1 < M, x − 1 < − , 1 < x < 1 − ,δ=− 1−x M M M 1 1 1 1 > M, 1 − x < , 1− < x < 1, δ = (b) if M > 0 and x < 1 then 1−x M M M

67. (a) if M < 0 and x > 1 then

1 1 1 1 > M, x < ,0<x< ,δ= x M M M 1 1 1 1 < x < 0, δ = − (b) if M < 0 and x < 0 then < M , −x < − , x M M M

68. (a) if M > 0 and x > 0 then

69. (a) Given any M > 0 there corresponds N > 0 such that if x > N then f (x) > M , x + 1 > M , x > M − 1, N = M − 1. (b) Given any M < 0 there corresponds N < 0 such that if x < N then f (x) < M , x + 1 < M , x < M − 1, N = M − 1.

2 70. (a) Given √ corresponds N > 0 such that if x > N then f (x) > M , x − 3 > M , √ any M > 0 there x > M + 3, N = M + 3. (b) Given any M < 0 there corresponds N < 0 such that if x < N then f (x) < M , x3 + 5 < M , x < (M − 5)1/3 , N = (M − 5)1/3 .

uh a

71. if δ ≤ 2 then |x − 3| < 2, −2 < x − 3 < 2, 1 < x < 5, and |x2 − 9| = |x + 3||x − 3| < 8|x − 3| < if |x − 3| < 18 , δ = min 2, 18

72. (a) We don’t care about the value of f at x = a, because the limit is only concerned with values of x near a. The condition that f be deﬁned for all x (except possibly x = a) is necessary, because if some points were excluded then the limit may not exist; for example, let f (x) = x if 1/x is not an integer and f (1/n) = 6. Then lim f (x) does not exist but it would if the points 1/n were excluded. √ (b) when x < 0 then x is not deﬁned

x→0

√

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x is deﬁned

Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Chapter 2

(b) no, x = 2 (f ) yes

(d) yes

2. (a) no, x = 2 (e) no, x = 2

(b) no, x = 2 (f ) yes

(d) yes

3. (a) no, x = 1, 3 (e) no, x = 3

(b) yes (f ) yes

(d) yes

4. (a) no, x = 3 (e) no, x = 3

(b) yes (f ) yes

us uf

1. (a) no, x = 2 (e) yes

(d) yes

5. (a) At x = 3 the one-sided limits fail to exist. (b) At x = −2 the two-sided limit exists but is not equal to F (−2). (c) At x = 3 the limit fails to exist.

6. (a) At x = 2 the two-sided limit fails to exist. (b) At x = 3 the two-sided limit exists but is not equal to F (3). (c) At x = 0 the two-sided limit fails to exist. 8. −2/5

(b) 3

7. (a) 3 y

x 3

(c)

(b)

9. (a)

1 x 1

(d)

-1

uh a

10. f (x) = 1/x, g(x) =

11. (a)

x 2

if x = 0

1 sin x

if x = 0

(b) One second could cost you one dollar.

t 1

EXERCISE SET 2.5

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13. none

14. none

15. none

16. f is not deﬁned at x = ±1 18. f is not deﬁned at x =

us uf

no; disasters (war, ﬂood, famine, pestilence, for example) can cause discontinuities continuous not usually continuous; see Exercise 11 continuous

17. f is not deﬁned at x = ±4

√ −7 ± 57 2

19. f is not deﬁned at x = ±3

20. f is not deﬁned at x = 0, −4

12. (a) (b) (c) (d)

Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Exercise Set 2.5

21. none

22. f is not deﬁned at x = 0, −3 16 is continuous on 4 < x; x lim f (x) = lim f (x) = f (4) = 11 so f is continuous at x = 4

23. none; f (x) = 2x + 3 is continuous on x < 4 and f (x) = 7 + x→4−

24.

x→4+

lim f (x) does not exist so f is discontinuous at x = 1

x→1

25. (a) f is continuous for x < 1, and for x > 1; lim− f (x) = 5, lim+ f (x) = k, so if k = 5 then f is x→1

x→1

continuous for all x

(b) f is continuous for x < 2, and for x > 2; lim− f (x) = 4k, lim+ f (x) = 4 + k, so if 4k = 4 + k, x→2

x→2

k = 4/3 then f is continuous for all x

(b) no, f is not deﬁned for x ≤ 2 (d) no, f is not deﬁned for x ≤ 2 y

(b)

27. (a)

26. (a) no, f is not deﬁned at x = 2 (c) yes

28. (a) f (c) = lim f (x) x→c

(b) lim f (x) = 2

lim g(x) = 1

uh a

x→1

-1

x→1

y 1

1 x

x 1

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Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Chapter 2

29. (a) x = 0, lim− f (x) = −1 = +1 = lim+ f (x) so the discontinuity is not removable x→0

x→0

(b) x = −3; deﬁne f (−3) = −3 = lim f (x), then the discontinuity is removable

x→−3

us uf

continuous there; at x = −2, lim does not exist, so the discontinuity is not removable x→−2

x→2

becomes continuous there

(c)

lim f (x) = 1 = 4 = lim f (x), so f has a nonremovable discontinuity at x = 2

x→2−

x→2+

lim f (x) = 8 = f (1), so f has a removable discontinuity at x = 1

x→1

31. (a) discontinuity at x = 1/2, not removable; at x = −3, removable

(b) 2x2 + 5x − 3 = (2x − 1)(x + 3)

(b)

x+2 1 1 = , so deﬁne f (2) = and f x2 + 2x + 4 3 3

30. (a) f is not deﬁned at x = 2; lim f (x) = lim

5 x 5

(b) one discontinuity at x = −1.52

32. (a) there appears to be one discontinuity near x = −1.52

-5

-3

–4

33. For x > 0, f (x) = x3/5 = (x3 )1/5 is the composition (Theorem 2.4.6) of the two continuous functions g(x) = x3 and h(x) = x1/5 and is thus continuous. For x < 0, f (x) = f (−x) which is the composition of the continuous functions f (x) (for positive x) and the continuous function y = −x. Hence f (−x) is continuous for all x > 0. At x = 0, f (0) = lim f (x) = 0. x→0

4 2 4 2 34. x √ + 7x + 1 ≥ 1 > 0, thus f (x) is the composition of the polynomial x + 7x + 1, the square root x, and the function 1/x and is therefore continuous by Theorem 2.5.6.

uh a

35. (a) Let f (x) = k for x = c and f (c) = 0; g(x) = l for x = c and g(c) = 0. If k = −l then f + g is continuous; otherwise it’s not. (b) f (x) = k for x = c, f (c) = 1; g(x) = l = 0 for x = c, g(c) = 1. If kl = 1, then f g is continuous; otherwise it’s not.

36. A rational function is the quotient f (x)/g(x) of two polynomials f (x) and g(x). By Theorem 2.5.2 f and g are continuous everywhere; by Theorem 2.5.3 f /g is continuous except when g(x) = 0.

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Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Exercise Set 2.5

(a) f (c) + g(c) = lim f (x) + lim g(x) = lim (f (x) + g(x)) so f + g is continuous at x = c. x→c

x→c

us uf

x→c

(b) same as (a) except the + sign becomes a − sign (c)

37. Since f and g are continuous at x = c we know that lim f (x) = f (c) and lim g(x) = g(c). In the x→c x→c following we use Theorem 2.2.2.

lim f (x) f (c) f (x) f = x→c = lim so is continuous at x = c g(c) lim g(x) x→c g(x) g x→c

38. h(x) = f (x) − g(x) satisﬁes h(a) > 0, h(b) < 0. Use the Intermediate Value Theorem or Theorem 2.5.9.

39. Of course such a function must be discontinuous. Let f (x) = 1 on 0 ≤ x < 1, and f (x) = −1 on 1 ≤ x ≤ 2.

40. A square whose diagonal has length r has area f (r) = r2 /2. Note that f (r) = r2 /2 < πr2 /2 < 2r2 = f (2r). By the Intermediate Value Theorem there must be a value c between r and 2r such that f (c) = πr2 /2, i.e. a square of diagonal c whose area is πr2 /2.

41. The cone has volume πr2 h/3. The function V (r) = πr2 h (for variable r and ﬁxed h) gives the volume of a right circular cylinder of height h and radius r, and satisﬁes V (0) < πr2 h/3 < V (r). By the Intermediate Value Theorem there is a value c between 0 and r such that V (c) = πr2 h/3, so the cylinder of radius c (and height h) has volume equal to that of the cone.

42. If f (x) = x3 − 4x + 1 then f (0) = 1, f (1) = −2. Use Theorem 2.5.9. 43. If f (x) = x3 + x2 − 2x then f (−1) = 2, f (1) = 0. Use the Intermediate Value Theorem. lim p(x) = −∞ and

x→−∞

lim p(x) = +∞ (or vice versa, if the leading coeﬃcient of p is

x→+∞

44. Since

negative), it follows that for M = −1 there corresponds N1 < 0, and for M = 1 there is N2 > 0, such that p(x) < −1 for x < N1 and p(x) > 1 for x > N2 . Choose x1 < N1 and x2 > N2 and use Theorem 2.5.9 on the interval [x1 , x2 ] to ﬁnd a solution of p(x) = 0.

45. For the negative root, use intervals on the x-axis as follows: [−2, −1]; since f (−1.3) < 0 and f (−1.2) > 0, the midpoint x = −1.25 of [−1.3, −1.2] is the required approximation of the root. For the positive root use the interval [0, 1]; since f (0.7) < 0 and f (0.8) > 0, the midpoint x = 0.75 of [0.7, 0.8] is the required approximation. 46. x = −1.25 and x = 0.75.

0.7

-2

uh a

0.8

-1

-5

-1

47. For the negative root, use intervals on the x-axis as follows: [−2, −1]; since f (−1.7) < 0 and f (−1.6) > 0, use the interval [−1.7, −1.6]. Since f (−1.61) < 0 and f (−1.60) > 0 the midpoint x = −1.605 of [−1.61, −1.60] is the required approximation of the root. For the positive root use the interval [1, 2]; since f (1.3) > 0 and f (1.4) < 0, use the interval [1.3, 1.4]. Since f (1.37) > 0 and f (1.38) < 0, the midpoint x = 1.375 of [1.37, 1.38] is the required approximation.

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Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Chapter 2

48. x = −1.605 and x = 1.375. 1

-1.7

us uf

-1.6 1.3

-0.5

-2

1.4

49. x = 2.24

a b + . Since lim+ f (x) = +∞ and lim− f (x) = −∞ there exist x1 > 1 and x−1 x−3 x→1 x→3 x2 < 3 (with x2 > x1 ) such that f (x) > 1 for 1 < x < x1 and f (x) < −1 for x2 < x < 3. Choose x3 in (1, x1 ) and x4 in (x2 , 3) and apply Theorem 2.5.9 on [x3 , x4 ].

50. Set f (x) =

51. The uncoated sphere has volume 4π(x − 1)3 /3 and the coated sphere has volume 4πx3 /3. If the volume of the uncoated sphere and of the coating itself are the same, then the coated sphere has twice the volume of the uncoated sphere. Thus 2(4π(x − 1)3 /3) = 4πx3 /3, or x3 − 6x2 + 6x − 2 = 0, with the solution x = 4.847 cm.

52. Let g(t) denote the altitude of the monk at time t measured in hours from noon of day one, and let f (t) denote the altitude of the monk at time t measured in hours from noon of day two. Then g(0) < f (0) and g(12) > f (12). Use Exercise 38. 53. We must show lim f (x) = f (c). Let > 0; then there exists δ > 0 such that if |x − c| < δ then x→c

|f (x) − f (c)| < . But this certainly satisﬁes Deﬁnition 2.4.1.

EXERCISE SET 2.6 1. none

4. x = nπ + π/2, n = 0, ±1, ±2, . . .

8. x = nπ + π/2, n = 0, ±1, ±2, . . .

9. 2nπ + π/6, 2nπ + 5π/6, n = 0, ±1, ±2, . . .

3 11. (a) sin √ x, x + 7x + 1 x, 3 + x, sin x, 2x (d)

12. (a) Use Theorem 2.5.6.

uh a

1 x→+∞ x lim

15. sin

5. x = nπ, n = 0, ±1, ±2, . . .

7. none

6. none

13. cos

3. x = nπ, n = 0, ±1, ±2, . . .

2. x = π

(b) |x|, sin x (e) sin x, sin x

= cos 0 = 1

1 , g(x) = x2 + 1 x2 + 1 2 14. sin lim = sin 0 = 0 x→+∞ x

(b) g(x) = cos x, g(x) =

πx x→+∞ 2 − 3x lim

10. none

√ π 3 = sin − =− 3 2

16.

sin h 1 1 lim = 2 h→0 h 2

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Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Exercise Set 2.6

20.

1 3

sin x lim x→0 x

θ→0+

2 =

1 θ

lim

θ→0+

1 3

6 sin 6x 8x 6 3 sin 6x sin 6x = , so lim = = x→0 sin 8x sin 8x 8 6x sin 8x 8 4

23.

7 sin 7x 3x 7 7 tan 7x tan 7x = so lim = (1)(1) = x→0 sin 3x 3 cos 7x 7x sin 3x sin 3x 3(1) 3

lim sin θ

θ→0

sin θ =0 θ→0 θ

25.

lim

√ sin x 1 lim x lim+ =0 5 x→0+ x x→0

21.

22.

24.

sin θ = +∞ θ

lim cos h

us uf

sin x 19. − lim− = −1 x x→0

lim

18.

h =1 h→0 sin h lim

h→0

sin h 1 + cos h sin h(1 + cos h) 1 + cos h sin h = = = ; no limit 1 − cos h 1 − cos h 1 + cos h 1 − cos2 h sin h

27.

θ2 1 + cos θ θ2 (1 + cos θ) = = 1 − cos θ 1 + cos θ 1 − cos2 θ

θ2 = (1)2 2 = 2 θ→0 1 − cos θ

(1 + cos θ) so lim

θ sin θ

26.

sin 3θ =3 θ→0 3θ

17. 3 lim

x 1 =1 x→0 cos 2π − x

28. cos( 12 π − x) = sin( 12 π) sin x = sin x, so lim

30.

t2 = 1 − cos2 t

29. 0

(1 − cos 5h)(1 + cos 5h)(1 + cos 7h) 1 − cos 5h 25 = = − cos 7h − 1 (cos 7h − 1)(1 + cos 5h)(1 + cos 7h) 49 1 − cos 5h 25 =− lim h→0 cos 7h − 1 49 1 = lim sin t; limit does not exist 32. lim+ sin t→+∞ x x→0

t sin t

sin 5h 5h

t2 =1 t→0 1 − cos2 t

, so lim

7h sin 7h

1 + cos 7h so 1 + cos 5h

31.

x→0

5.1 0.098845

sin x = −3 x

36.

lim x − 3 lim

x→0

5.01 0.099898

34.

1 33. lim cos = lim cos t; limit does not exist + t→+∞ x x→0

5.001 0.99990

5.0001 0.099999

35. 2 + lim

x→0

5.00001 0.100000

4.9 0.10084

sin x =3 x

4.99 0.10010

4.999 0.10001

4.9999 0.10000

4.99999 0.10000

The limit is 0.1. 2.1 0.484559

uh a

37.

2.01 0.498720

2.001 0.499875

2.0001 0.499987

2.00001 0.499999

1.9 0.509409

1.99 0.501220

1.999 0.500125

1.9999 0.500012

1.99999 0.500001

The limit is 0.5.

38.

−1.9 −1.99 −1.999 −1.9999 −1.99999 −2.1 −2.01 −2.001 −2.0001 −2.00001 −0.898785 −0.989984 −0.999000 −0.999900 −0.999990 −1.097783 −1.009983 −1.001000 −1.000100 −1.000010

The limit is −1.

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39.

Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Chapter 2

−0.9 0.405086

−0.99 0.340050

−0.999 0.334001

−0.9999 0.333400

−0.99999 0.333340

−1.1 0.271536

−1.01 0.326717

−1.001 0.332667

−1.0001 0.333267

−1.00001 0.333327

40. k = f (0) = lim

x→0

41.

sin 3x sin 3x = 3 lim = 3, so k = 3 x→0 x 3x

lim f (x) = k lim

x→0−

us uf

The limit is 1/3.

x→0

sin kx 1 = k, lim+ f (x) = 2k 2 , so k = 2k 2 , k = kx cos kx 2 x→0

43. (a)

lim

t→0+

sin t =1 t

(b)

x→π

cos(π/x) (π − 2t) sin t π − 2t sin t π − t = sin t, so lim = lim = lim lim = x→2 x − 2 t→0 t→0 2 4t 4 t→0 t 4

45. t = x − 1; sin(πx) = sin(πt + π) = − sin πt; and lim

x→1

46. t = x − π/4; tan x − 1 =

≤ |x|

48. −x ≤ x sin 2

lim f (x) = 1 by the Squeezing Theorem

x→0

y = cos x 0

≤ x2

lim f (x) = 0 by the Squeezing Theorem

x→+∞

x 4

uh a

-1

sin x 1 1 and h(x) = ; thus lim = 0 by the Squeezing Theorem. x→+∞ x x x

51. Let g(x) = − y

-1

y = 1 – x2

52.

50π √ 3 x

-1

y = f (x)

50.

y 1

47. −|x| ≤ x cos

50π x

sin(πx) sin πt = − lim = −π t→0 x−1 t

2 sin t tan x − 1 2 sin t ; lim = lim =2 t→0 t(cos t − sin t) cos t − sin t x→π/4 x − π/4

49.

1 − cos t = 0 (Theorem 2.6.3) t

44. cos

π−x t =1 = lim t→0 sin t sin x

lim

t→0−

42. No; sin x/|x| has unequal one-sided limits.

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Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Exercise Set 2.6

πx 54. cos x = cos t where x is measured in degrees, t in radians, and t = . Thus 180 1 − cos x 1 − cos t = lim = 0. lim x→0 t→0 (180t/π) x

us uf

πx . Thus 53. (a) sin x = sin t where x is measured in degrees, t is measured in radians and t = 180 π sin x sin t lim = lim = . x→0 x t→0 (180t/π) 180

π π ≈ = 0.17453 18 18

(b) sin 10◦ = sin

56. (a) cos θ = cos 2α = 1 − 2 sin2 (θ/2)

(b) cos 10◦ = 0.98481

55. (a) sin 10◦ = 0.17365

≈ 1 − 2(θ/2)2 = 1 − 12 θ2 1 π 2 ≈ 0.98477 (c) cos 10◦ = 1 − 2 18 (b) tan 5◦ ≈

57. (a) 0.08749

(b) Since α is small, tan α◦ ≈ (c) h ≈ 52.36 ft

58. (a) h = 52.55 ft

π = 0.08727 36

πα is a good approximation. 180

(b)

59. (a) Let f (x) = x − cos x; f (0) = −1, f (π/2) = π/2. By the IVT there must be a solution of f (x) = 0. y

1.5

y=x

0.5

1 y = cos x

c/2

(b)

60. (a) f (x) = x + sin x − 1; f (0) = −1, f (π/6) = π/6 − 1/2 > 0. By the IVT there must be a solution of f (x) = 0. y = 1 – sin x

0.5

y=x

c/6

uh a

61. (a) There is symmetry about the equatorial plane. (b) Let g(φ) be the given function. Then g(38) < 9.8 and g(39) > 9.8, so by the Intermediate Value Theorem there is a value c between 38 and 39 for which g(c) = 9.8 exactly.

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Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Chapter 2

us uf

62. (a) does not exist (b) the limit is zero (c) For part (a) consider the fact that given any δ > 0 there are inﬁnitely many rational numbers x satisfying |x| < δ and there are inﬁnitely many irrational numbers satisfying the same condition. Thus if the limit were to exist, it could not be zero because of the rational numbers, and it could not be 1 because of the irrational numbers, and it could not be anything else because of all the numbers. Hence the limit cannot exist. For part (b) use the Squeezing Theorem with +x and −x as the ‘squeezers’.

(b) no limit (e) 3 (h) 2

2. (a) f (x) = 2x/(x − 1)

y 10

1. (a) 1 (d) 1 (g) 0

CHAPTER 2 SUPPLEMENTARY EXERCISES

a+b a+b and f (x) = 1 for ≤x≤b 2 2

4. f (x) = −1 for a ≤ x <

5. (a) 0.222 . . . , 0.24390, 0.24938, 0.24994, 0.24999, 0.25000; for x = 2, f (x) = so the limit is 1/4.

1 , x+2

(b) 1.15782, 4.22793, 4.00213, 4.00002, 4.00000, 4.00000; to prove, sin 4x 4 sin 4x tan 4x = , the limit is 4. = use x x cos 4x cos 4x 4x

7. (a)

x f (x)

(b) none

6. (a) y = 0 1 1.000

0.1 0.443

0.01 0.409

0.001 0.406

0.0001 0.406

0.000001 0.405

(b)

uh a

0.5

-1

8. (a) 0.4 amperes (d) 0.0187

x 1

(b) [0.3947, 0.4054] (e)

(c)

3 3 , 7.5 + δ 7.5 − δ

It becomes inﬁnite.

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Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Chapter 2 Supplementary Exercises

y 1

us uf

9. (a)

0.4 x 0.2

0.8

(b) Let g(x) = x − f (x). Then g(1) ≥ 0 and g(0) ≤ 0; by the Intermediate Value Theorem there is a solution c in [0, 1] of g(c) = 0. 1 − cos θ 1 − cos2 θ 1 − cos θ = tan lim = tan 0 = 0 = tan lim θ→0 θ→0 θ(1 + cos θ) θ→0 θ θ √ √ √ t−1 t−1 t−1 t+1 (t − 1)( t + 1) √ √ (b) √ =√ = = lim ( t + 1) = 2 = t + 1; lim √ t→1 t−1 t−1 t−1 t+1 t − 1 t→1

lim tan

10. (a)

(2 − 1/x)5 (2x − 1)5 → 25 /3 = 32/3 as x → +∞ = 3 + 2x − 7)(x − 9x) (3 + 2/x − 7/x2 )(1 − 9/x2 ) sin(θ + π) − sin θ = lim cos (d) sin(θ + π) = sin θ cos π − cos θ sin π = − sin θ, so lim cos θ→0 θ→0 2θ 2θ 1 − sin θ = cos lim = cos − θ→0 2θ 2 (3x2

(c)

11. If, on the contrary, f (x0 ) < 0 for some x0 in [0, 1], then by the Intermediate Value Theorem we would have a solution of f (x) = 0 in [0, x0 ], contrary to the hypothesis.

x→2

12. For x < 2 f is a polynomial and is continuous; for x > 2 f is a polynomial and is continuous. At x = 2, f (2) = −13 = 13 = lim+ f (x) so f is not continuous there.

13. f (−6) = 185, f (0) = −1, f (2) = 65; apply Theorem 2.4.9 twice, once on [−6, 0] and once on [0, 2] 14. 3.317

y 1

16.

15. Let = f (x0 )/2 > 0; then there corresponds δ > 0 such that if |x − x0 | < δ then |f (x) − f (x0 )| < , − < f (x) − f (x0 ) < , f (x) > f (x0 ) − = f (x0 )/2 > 0 for x0 − δ < x < x0 + δ.

uh a

17. (a) −3.449, 1.449

(b) x = 0, ±1.896

18. Since lim sin(1/x) does not exist, no conclusions can be drawn. x→0

√ √ √ 19. (a) 5, no limit, 10, 10, no limit, +∞, no limit (b) 5, 10, 0, 0, 10, −∞, +∞

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Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Chapter 2

21. a/b

(b) −1, +1, −1, −1, no limit, −1, +1

22. 1

24. 2

23. does not exist 27. 3 − k

26. k 2

25. 0

20. (a) −1/5, +∞, −1/10, −1/10, no limit, 0, 0

us uf

x f (x)

1 0.236

0.1 0.2498

0.01 0.2500

0.001 0.2500

0.0001 0.25000

(b)

√ x2 + 4 − 2 x2 + 4 + 2 x2 1 √ √ = =√ , so 2 x x2 + 4 + 2 x2 ( x2 + 4 + 2) x2 + 4 + 2 √ x2 + 4 − 2 1 1 lim = lim √ = x→0 x→0 4 x2 x2 + 4 + 2 0.00001 0.00000

√ 30. (a)

28. The numerator satisﬁes: |2x + x sin 3x| ≤ |2x| + |x| = 3|x|. Since the denominator grows like x2 , the limit is 0.

The division may entail division by zero (e.g. on an HP 42S), or the numerator may be inaccurate (catastrophic subtraction, e.g.).

0.1 2.59

0.01 2.70

0.001 2.717

0.0001 2.718

32.

x f (x)

3.1 5.74

3.01 5.56

3.001 5.547

3.0001 5.545

33.

x f (x)

1.1 0.49

1.01 0.54

1.001 0.540

34.

x f (x)

0.1 99.0

0.01 9048.8

35.

x f (x)

1.0001 0.5403

uh a

100 0.48809

0.001 368063.3

1000 0.49611

0.00001 2.7183

3.00001 5.5452

3.000001 5.54518

1.00001 0.54030

1.000001 0.54030

0.0001 4562.7

104 0.49876

0.000001 2.71828

x f (x)

31.

0.00001 3.9 × 10−34

105 0.49961

106 0.49988

0.000001 0

107 0.49996

36. For large values of x (not much more than 100) the computer can’t handle 5x or 3x , yet the limit is 5. 38. $2,001.60, $2,009.66, $2,013.62, $2013.75

37. δ ≈ 0.07747 (use a graphing utility)

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Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Chapter 2 Supplementary Exercises

39. (a) x3 − x − 1 = 0, x3 = x + 1, x =

√ 3

x + 1.

(b)

x -1

1 -1

(d) 1, 1.26, 1.31, 1.322, 1.324, 1.3246, 1.3247

(c)

us uf

x1 x2 x3

(b) 0, −1, −2, −9, −730

40. (a) 20

√ 5

x + 2; 1.267168

42. x = cos x; 0.739085 (after 33 iterations!).

uh a

41. x =

x -1

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Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH CHAPTER 3

EXERCISE SET 3.1 1. (a) mtan = (50 − 10)/(15 − 5) = 40/10 = 4 m/s

(b)

v (m/s) 4

(10 − 10)/(3 − 0) = 0 cm/s t = 0, t = 2, and t = 4.2 (horizontal tangent line) maximum: t = 1 (slope > 0) minimum: t = 3 (slope < 0) (3 − 18)/(4 − 2) = −7.5 cm/s (slope of estimated tangent line to curve at t = 3)

2. (a) (b) (c) (d)

t (s)

us uf

The Derivative

3. From the ﬁgure:

t t0 t1

t t1

(a) The particle is moving faster at time t0 because the slope of the tangent to the curve at t0 is greater than that at t2 . (b) The initial velocity is 0 because the slope of a horizontal line is 0. (c) The particle is speeding up because the slope increases as t increases from t0 to t1 . (d) The particle is slowing down because the slope decreases as t increases from t1 to t2 .

5. It is a straight line with slope equal to the velocity.

decreasing (slope of tangent line decreases with increasing time) increasing (slope of tangent line increases with increasing time) increasing (slope of tangent line increases with increasing time) decreasing (slope of tangent line decreases with increasing time)

6. (a) (b) (c) (d)

f (4) − f (3) (4)2 /2 − (3)2 /2 7 = = 4−3 1 2 f (x1 ) − f (3) x21 /2 − 9/2 = lim = lim x1 →3 x1 →3 x1 − 3 x1 − 3

uh a

7. (a) msec = mtan

x21 − 9 (x1 + 3)(x1 − 3) x1 + 3 = lim = lim =3 x1 →3 2(x1 − 3) x1 →3 x →3 2(x1 − 3) 2 1

= lim

(b)

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Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Exercise Set 3.1

mtan = lim

x1 →x0

f (x1 ) − f (x0 ) x1 − x0

(d)

y Tangent

x2 /2 − x20 /2 = lim 1 x1 →x0 x1 − x0 x21 − x20 x1 →x0 2(x1 − x0 ) x1 + x0 = lim = x0 x1 →x0 2

us uf

(c)

= lim

Secant

mtan

23 − 13 f (2) − f (1) = =7 2−1 1 f (x1 ) − f (1) x3 − 1 (x1 − 1)(x21 + x1 + 1) = lim = lim = lim 1 x1 →1 x1 →1 x1 − 1 x1 →1 x1 − 1 x1 − 1

8. (a) msec = (b)

= lim (x21 + x1 + 1) = 3

= lim

x1 →x0

x31 − x30 x1 − x0

= lim (x21 + x1 x0 + x20 ) x1 →x0

Secant

x1 →2

= lim

2 − x1 −1 1 =− = lim 2x1 (x1 − 2) x1 →2 2x1 4

mtan = lim

x1 →x0

f (x1 ) − f (x0 ) x1 − x0

(d)

(c)

Tangent

f (3) − f (2) 1/3 − 1/2 1 = =− 3−2 1 6 f (x1 ) − f (2) 1/x1 − 1/2 = lim = lim x1 →2 x1 →2 x1 − 2 x1 − 2

9. (a) msec = mtan

= 3x20

(b)

(d)

mtan

f (x1 ) − f (x0 ) = lim x1 →x0 x1 − x0

x1 →1

(c)

1/x1 − 1/x0 = lim x1 →x0 x1 − x0 x0 − x1 = lim x1 →x0 x0 x1 (x1 − x0 )

Secant

x 1

Tangent

uh a

−1 1 = lim =− 2 x1 →x0 x0 x1 x0

10. (a) msec =

(b)

1/4 − 1 3 f (2) − f (1) = =− 2−1 1 4 f (x1 ) − f (1) 1/x21 − 1 = lim x1 →1 x1 →1 x1 − 1 x1 − 1

mtan = lim

= lim

1 − x21 −(x1 + 1) = −2 = lim − 1) x1 →1 x21

x1 →1 x2 1 (x1

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Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Chapter 3

x1 →x0

= lim

1/x21 − 1/x20 x1 − x0

= lim

x20 − x21 2 x0 x21 (x1 − x0 )

x1 →x0

= lim

x1 →x0

11. (a)

f (x1 ) − f (x0 ) x1 − x0

mtan = lim

x1 →x0

= lim

x1 →x0

1 x 2

−(x1 + x0 ) 2 =− 3 x20 x21 x0

Tangent

f (x1 ) − f (x0 ) (x21 + 1) − (x20 + 1) = lim x1 →x0 x1 − x0 x1 − x0 x21 − x20 = lim (x1 + x0 ) = 2x0 x1 →x0 x1 − x0

f (x1 ) − f (x0 ) (x21 + 3x1 + 2) − (x20 + 3x0 + 2) = lim x1 →x0 x1 − x0 x1 − x0

= lim

(x21 − x20 ) + 3(x1 − x0 ) = lim (x1 + x0 + 3) = 2x0 + 3 x1 →x0 x1 − x0

x1 →x0

mtan = lim

x1 →x0

(b) mtan = 2(2) + 3 = 7

x1 →x0

1 1 (b) mtan = √ = 2 2 1

√ √ 1/ x1 − 1/ x0 f (x1 ) − f (x0 ) = lim x1 →x0 x1 →x0 x1 − x0 x1 − x0 √ √ x0 − x1 −1 1 = lim √ √ = lim √ √ = − 3/2 √ √ x1 →x0 x →x x0 x1 ( x1 + x0 ) x0 x1 (x1 − x0 ) 1 0 2x0

mtan = lim

1 1 =− 3/2 16 2(4)

(b) mtan = −

14. (a)

1 1 √ = √ x1 + x0 2 x0

√ x1 − x0 x1 − x0

= lim √

√

mtan

f (x1 ) − f (x0 ) = lim = lim x1 →x0 x1 →x0 x1 − x0

13. (a)

Secant

(b) mtan = 2(2) = 4 12. (a)

(d)

mtan = lim

us uf

(c)

15. (a) 72◦ F at about 4:30 P.M. (b) about (67 − 43)/6 = 4◦ F/h (c) decreasing most rapidly at about 9 P.M.; rate of change of temperature is about −7◦ F/h (slope of estimated tangent line to curve at 9 P.M.)

uh a

16. For V = 10 the slope of the tangent line is about −0.25 atm/L, for V = 25 the slope is about −0.04 atm/L.

17. (a) during the ﬁrst year after birth (b) about 6 cm/year (slope of estimated tangent line at age 5) (c) the growth rate is greatest at about age 14; about 10 cm/year

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Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Exercise Set 3.2

(d) 40 Growth rate (cm/year)

us uf

20 10 t (yrs) 5

18. (a) The rock will hit the ground when 16t2 = 576, t2 = 36, t = 6 s (only t ≥ 0 is meaningful) 16(6)2 − 16(0)2 = 96 ft/s 6−0

16(3)2 − 16(0)2 = 48 ft/s 3−0

(b) vave =

16t21 − 16(6)2 16(t21 − 36) = lim t1 →6 t1 →6 t1 − 6 t1 − 6

(d) vinst = lim

= lim 16(t1 + 6) = 192 ft/s

t1 →6

19. (a) 5(40)3 = 320,000 ft

(b) vave = 320,000/40 = 8,000 ft/s

(c) 5t = 135 when the rocket has gone 135 ft, so t3 = 27, t = 3 s; vave = 135/3 = 45 ft/s. 5t31 − 5(40)3 5(t31 − 403 ) = lim t1 →40 t1 →40 t1 − 40 t1 − 40

(d) vinst = lim

= lim 5(t21 + 40t1 + 1600) = 24, 000 ft/s [3(3)2 + 3] − [3(1)2 + 1] = 13 mi/h 3−1

20. (a) vave =

t1 →40

(3t21 + t1 ) − 4 (3t1 + 4)(t1 − 1) = lim = lim (3t1 + 4) = 7 mi/h t1 →1 t1 →1 t1 →1 t1 − 1 t1 − 1

21. (a) vave =

(b) vinst = lim

6(4)4 − 6(2)4 = 720 ft/min 4−2

6t41 − 6(2)4 6(t41 − 16) = lim t1 →2 t1 →2 t1 − 2 t1 − 2

(b) vinst = lim

6(t21 + 4)(t21 − 4) = lim 6(t21 + 4)(t1 + 2) = 192 ft/min t1 →2 t1 →2 t1 − 2

= lim

EXERCISE SET 3.2

1. f (1) = 2, f (3) = 0, f (5) = −2, f (6) = −1/2

uh a

2. f (4) < f (0) < f (2) < 0 < f (−3)

3. (b) m = f (2) = 3 4−3 =1 0 − (−1)

4. f (−1) = m =

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Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Chapter 3

8. y − 3 = −4(x + 2), y = −4x − 5

7. y − (−1) = 5(x − 3), y = 5x − 16

us uf

f (w) − f (x) 3w2 − 3x2 = lim = lim 3(w + x) = 6x; f (3) = 3(3)2 = 27, f (3) = 18 w→x w→x w→x w−x w−x so y − 27 = 18(x − 3), y = 18x − 27

9. f (x) = lim

f (w) − f (x) w4 − x4 = lim = lim (w3 + w2 x + wx2 + x3 ) = 4x3 ; w→x w→x w − x w→x w−x f (−2) = (−2)4 = 16, f (−2) = −32 so y − 16 = −32(x + 2), y = −32x − 48 f (x) = lim

10.

f (w) − f (x) w3 − x3 = lim = lim (w2 + wx + x2 ) = 3x2 ; f (0) = 03 = 0, w→x w − x w→x w→x w−x f (0) = 0 so y − 0 = (0)(x − 0), y = 0

11. f (x) = lim

f (w) − f (x) 2w3 + 1 − (2x3 + 1) = lim = lim 2(w2 + wx + x2 ) = 6x2 ; w→x w→x w→x w−x w−x f (−1) = 2(−1)3 + 1 = −1, f (−1) = 6 so y + 1 = 6(x + 1), y = 6x + 5 lim

12. f (x) =

√ √ f (w) − f (x) w+1− x+1 = lim 13. f (x) = lim w→x w→x w−x w−x √ √ √ √ w+1− x+1 w+1+ x+1 1 1 = √ √ √ √ = lim = lim √ ; w→x w→x w−x w+1+ x+1 w+1+ x+1 2 x+1 √ 1 1 1 5 f (8) = 8 + 1 = 3, f (8) = so y − 3 = (x − 8), y = x + 6 6 6 3 √ √ f (w) − f (x) 2w + 1 − 2x + 1 14. f (x) = lim = lim w→x w→x w−x w−x 2 2 1 √ = lim √ =√ = lim √ w→x w→x 2w + 1 + 2x + 1 2x + 1 9 + 2h + 3 √ 1 1 5 f (4) = 2(4) + 1 = 9 = 3, f (4) = 1/3 so y − 3 = (x − 4), y = x + 3 3 3

x − (x + ∆x) 1 1 − x(x + ∆x) 15. f (x) = lim x + ∆x x = lim ∆x→0 ∆x→0 ∆x ∆x

uh a

= lim

−∆x 1 1 = lim − =− 2 x∆x(x + ∆x) ∆x→0 x(x + ∆x) x

∆x→0

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Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Exercise Set 3.2

x + 1 − x − ∆x − 1 −∆x = lim ∆x(x + 1)(x + ∆x + 1) ∆x→0 ∆x(x + 1)(x + ∆x + 1)

= lim

1 −1 =− (x + 1)(x + ∆x + 1) (x + 1)2

∆x→0

us uf

= lim

1 1 (x + 1) − (x + ∆x + 1) − (x + ∆x) + 1 x + 1 (x + 1)(x + ∆x + 1) = lim 16. f (x) = lim ∆x→0 ∆x→0 ∆x ∆x

[a(x + ∆x)2 + b] − [ax2 + b] ax2 + 2ax∆x + a(∆x)2 + b − ax2 − b = lim ∆x→0 ∆x→0 ∆x ∆x

17. f (x) = lim

2ax∆x + a(∆x)2 = lim (2ax + a∆x) = 2ax ∆x→0 ∆x→0 ∆x

= lim

2x∆x + ∆x2 − ∆x (x + ∆x)2 − (x + ∆x) − (x2 − x) = lim ∆x→0 ∆x→0 ∆x ∆x

= lim (2x − 1 + ∆x) = 2x − 1 ∆x→0

∆x→0

= lim

−1 1 x − (x + ∆x) √ √ = lim √ √ = − 3/2 √ √ √ √ 2x ∆x x x + ∆x( x + x + ∆x) ∆x→0 x x + ∆x( x + x + ∆x)

∆x→0

1 1 √ √ −√ x − x + ∆x x x + ∆x = lim √ √ ∆x→0 ∆x x x + ∆x ∆x

19. f (x) = lim

√

18. f (x) = lim

1 x2 − (x + ∆x)2 1 − (x + ∆x)2 x2 x2 (x + ∆x)2 = lim 20. f (x) = lim ∆x→0 ∆x→0 ∆x ∆x

x2 − x2 − 2x∆x − ∆x2 −2x∆x − ∆x2 −2x − ∆x 2 = lim = lim 2 =− 3 2 2 2 2 2 ∆x→0 ∆x→0 x ∆x(x + ∆x) ∆x→0 x (x + ∆x) x ∆x(x + ∆x) x

= lim

f (t + h) − f (t) [4(t + h)2 + (t + h)] − [4t2 + t] = lim h→0 h→0 h h

21. f (t) = lim

4t2 + 8th + 4h2 + t + h − 4t2 − t h→0 h

= lim

4 4 4 π(r + h)3 − πr3 π(r3 + 3r2 h + 3rh2 + h3 − r3 ) dV 3 3 3 = lim = lim h→0 h→0 dr h h 4 π(3r2 + 3rh + h2 ) = 4πr2 3

22.

8th + 4h2 + h = lim (8t + 4h + 1) = 8t + 1 h→0 h→0 h

= lim

h→0

(b) F

(d) C

(e) A

(f ) E

uh a

23. (a) D

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Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Chapter 3

24. Any function of the form f (x) = x + k has slope 1, and thus the derivative must be equal to 1 everywhere.

us uf

y 3

x 1

-2

25. (a)

(b)

(c)

-2

x x 1

-1

(b)

26. (a)

(c)

√

(b) f (x) =

28. (a) f (x) = x7 and a = 1

(b) f (x) = cos x and a = π

dy [4(x + h)2 + 1] − [4x2 + 1] 4x2 + 8xh + 4h2 + 1 − 4x2 − 1 = lim = lim = lim (8x + 4h) = 8x h→0 h→0 dx h→0 h h dy = 8(1) = 8 dx x=1

uh a

29.

x and a = 1

27. (a) f (x) = x2 and a = 3

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Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Exercise Set 3.2

5 5 5x − 5(x + h) 5 5 +1 − +1 − dy x+h x x(x + h) = lim = lim x + h x = lim h→0 h→0 dx h→0 h h h 5x − 5x − 5h −5 5 = lim = lim =− 2 h→0 hx(x + h) h→0 x(x + h) x dy 5 5 =− =− 2 dx x=−2 (−2) 4

31. y = −2x + 1

1.5

32. 5

-2

us uf

30.

2 2.5

0 0

33. (b)

h (f (1 + h) − f (1))/h

0.5 0.1 1.6569 1.4355

34. (b)

h (f (1 + h) − f (1))/h

0.5 0.50489

-3

0.01 0.001 0.0001 0.00001 1.3911 1.3868 1.3863 1.3863

0.1 0.01 0.67060 0.70356

0.001 0.0001 0.70675 0.70707

0.00001 0.70710

35. (a) dollars/ft (b) As you go deeper the price per foot may increase dramatically, so f (x) is roughly the price per additional foot. (c) If each additional foot costs extra money (this is to be expected) then f (x) remains positive. f (301) − f (300) (d) From the approximation 1000 = f (300) ≈ 301 − 300 we see that f (301) ≈ f (300) + 1000, so the extra foot will cost around $1000.

36. (a) gallons/dollar (b) The increase in the amount of paint that would be sold for one extra dollar. (c) It should be negative since an increase in the price of paint would decrease the amount of paint sold. f (11) − f (10) (d) From −100 = f (10) ≈ we see that f (11) ≈ f (10) − 100, so an increase of one 11 − 10 dollar would decrease the amount of paint sold by around 100 gallons. 37. (a) F ≈ 200 lb, dF/dθ ≈ 50 lb/rad

(b) µ = (dF/dθ)/F ≈ 50/200 = 0.25

10 − 2.2 = 0.078 billion, or in 2050 the world population 2050 − 1950 was increasing at the rate of about 78 million per year. dN 0.078 (b) =≈ = 0.013 = 1.3 %/year dt 6

uh a

38. (a) The slope of the tangent line ≈

39. (a) T ≈ 115◦ F, dT /dt ≈ −3.35◦ F/min (b) k = (dT /dt)/(T − T0 ) ≈ (−3.35)/(115 − 75) = −0.084

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Chapter 3

lim f (x) = lim

√ 3

x = 0 = f (0), so f is continuous at x = 0. √ 3 f (0 + h) − f (0) h−0 1 = lim = lim 2/3 = +∞, so lim h→0 h→0 h→0 h h h x→0

x→0

-2

f (0) does not exist.

42.

lim f (x) = lim (x − 2)2/3 = 0 = f (2) so f is continuous at

x→2

h2/3 − 0 1 f (2 + h) − f (2) = lim = lim 1/3 h→0 h h→0 h→0 h h

41.

Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

us uf

x = 2. lim

which does not exist so f (2) does not exist.

lim f (x) = lim+ f (x) = f (1), so f is continuous at x = 1.

x→1−

x→1

lim

[(1 + h)2 + 1] − 2 f (1 + h) − f (1) = lim = lim− (2 + h) = 2; h h h→0− h→0

lim

f (1 + h) − f (1) 2(1 + h) − 2 = lim+ = lim+ 2 = 2, so f (1) = 2. h h h→0 h→0

h→0−

-3

lim f (x) = lim f (x) = f (1) so f is continuous at x = 1.

x→1−

x→1+

lim

f (1 + h) − f (1) [(1 + h)2 + 2] − 3 = lim = lim (2 + h) = 2; h h h→0− h→0−

lim

f (1 + h) − f (1) [(1 + h) + 2] − 3 = lim = lim+ 1 = 1, + h h h→0 h→0

h→0−

h→0+

x -3

so f (1) does not exist.

44.

h→0+

43.

45. Since −|x| ≤ x sin(1/x) ≤ |x| it follows by the Squeezing Theorem

(Theorem 2.6.2) that lim x sin(1/x) = 0. The derivative cannot x→0

f (x) − f (0) = sin(1/x). This function oscillates x between −1 and +1 and does not tend to zero as x tends to zero.

uh a

exist: consider

46. For continuity, compare with ±x2 to establish that the limit is zero. The diﬀerential quotient is x sin(1/x) and (see Exercise 45) this has a limit of zero at the origin.

47. f is continuous at x = 1 because it is diﬀerentiable there, thus lim f (1 + h) = f (1) and so f (1) = 0 h→0

f (1 + h) f (1 + h) − f (1) f (1 + h) exists; f (1) = lim = lim = 5. because lim h→0 h→0 h→0 h h h

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Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Exercise Set 3.3

f (x + h) − f (x) , but h→0 h

48. Let x = y = 0 to get f (0) = f (0) + f (0) + 0 so f (0) = 0. f (x) = lim

us uf

f (x + h) = f (x) + f (h) + 5xh so f (x + h) − f (x) = f (h) + 5xh and f (h) f (h) + 5xh = lim + 5x = 3 + 5x. f (x) = lim h→0 h→0 h h f (x + h) − f (x) f (x)f (h) − f (x) f (x)[f (h) − 1] = lim = lim h→0 h→0 h→0 h h h

f (x) = lim

= f (x) lim

h→0

f (h) − f (0) = f (x)f (0) = f (x) h

49.

1. 28x6

2. −36x11

5. 0

9. 3ax + 2bx + c

10.

1 a

1 2x + b

13. −3x−4 − 7x−8

√ 11. 24x−9 + 1/ x

1 2x − 4

1 + 2x − 4

5 12. −42x−7 − √ 2 x

1 1 √ − 2 x x2

d 1 2 2 (3x + 6) = (3x + 6)(2) + 2x − (6x) dx 4

3 = 18x2 − x + 12 2 16. f (x) = (2 − x − 3x3 )

2 x 5

14.

d 15. f (x) = (3x + 6) dx

4. 2x3

1 7. − (7x6 + 2) 3

√

3. 24x7 + 2

EXERCISE SET 3.3

d d (7 + x5 ) + (7 + x5 ) (2 − x − 3x3 ) dx dx

= (2 − x − 3x3 )(5x4 ) + (7 + x5 )(−1 − 9x2 )

= −24x7 − 6x5 + 10x4 − 63x2 − 7 17. f (x) = (x3 + 7x2 − 8)

d d (2x−3 + x−4 ) + (2x−3 + x−4 ) (x3 + 7x2 − 8) dx dx

= (x3 + 7x2 − 8)(−6x−4 − 4x−5 ) + (2x−3 + x−4 )(3x2 + 14x)

= −15x−2 − 14x−3 + 48x−4 + 32x−5 18. f (x) = (x−1 + x−2 )

d d (3x3 + 27) + (3x3 + 27) (x−1 + x−2 ) dx dx

uh a

= (x−1 + x−2 )(9x2 ) + (3x3 + 27)(−x−2 − 2x−3 ) = 3 + 6x − 27x−2 − 54x−3

19. 12x(3x2 + 1)

21.

20.

f (x) = x10 + 4x6 + 4x2 , f (x) = 10x9 + 24x5 + 8x

d d (5x − 3) (1) − (1) (5x − 3) 5 dy dx dx =− ; y (1) = −5/4 = 2 dx (5x − 3)2 (5x − 3)

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25.

d d (3t) (t2 + 1) − (t2 + 1) (3t) dx (3t)(2t) − (t2 + 1)(3) t2 − 1 dt dt = = = 9t2 3t2 dt (3t)2 d d (x + 3) (2x − 1) − (2x − 1) (x + 3) dy dx dx = dx (x + 3)2 =

26.

7 dy 7 (x + 3)(2) − (2x − 1)(1) = ; = 2 2 (x + 3) (x + 3) dx x=1 16

d d (x2 − 5) (4x + 1) − (4x + 1) (x2 − 5) dy dx dx = dx (x2 − 5)2

d d (2t + 1) (3t) − (3t) (2t + 1) (2t + 1)(3) − (3t)(2) 3 dx dt dt = = = dt (2t + 1)2 (2t + 1)2 (2t + 1)2

us uf

24.

√ d √ d ( x + 2) (3) − 3 ( x + 2) √ √ dy dx√ dx = = −3/(2 x( x + 2)2 ); y (1) = −3/18 = −1/6 dx ( x + 2)2

23.

Chapter 3

22.

Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

(x2 − 5)(4) − (4x + 1)(2x) 4x2 + 2x + 20 dy 13 = =− ; = 2 2 2 2 (x − 5) dx x=1 8 (x − 5) 3x + 2 x

27.

−5 d 3x + 2 d −5 x +1 + x +1 dx dx x x(3) − (3x + 2)(1) 3x + 2 = −5x−6 + x−5 + 1 x x2 3x + 2 2 −5x−6 + x−5 + 1 − 2 ; = x x

dy = dx

dy = 5(−5) + 2(−2) = −29 dx x=1

x−1 d dy d x−1 = (2x7 − x2 ) + (2x7 − x2 ) dx dx x + 1 x + 1 dx x−1 (x + 1)(1) − (x − 1)(1) (14x6 − 2x) = (2x7 − x2 ) + (x + 1)2 x+1 2 x−1 (14x6 − 2x); = (2x7 − x2 ) · + (x + 1)2 x+1 dy 1 2 = (2 − 1) + 0(14 − 2) = dx x=1 4 2

28.

0.999699 − (−1) f (1.01) − f (1) = = 0.0301, and by diﬀerentiation, f (1) = 3(1)2 − 3 = 0 0.01 0.01

uh a 29. f (1) ≈

1.01504 − 1 f (1.01) − f (1) = = 1.504, and by diﬀerentiation, 0.01 0.01 √ x f (1) = x+ √ = 1.5 2 x x=1

30. f (1) ≈

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31. f (1) = 0

32. f (1) = 1

33. 32t

34. 2π

35. 3πr2

36. −2α−2 + 1

1 d 1 λ0 + 6λ5 d λλ0 + λ6 (λλ0 + λ6 ) = = (λ0 + 6λ5 ) = dλ 2 − λ0 2 − λ0 dλ 2 − λ0 2 − λ0

39. (a) g (x) = (b) g (x) =

√

1 1 xf (x) + √ f (x), g (4) = (2)(−5) + (3) = −37/4 4 2 x

(4)(−5) − 3 xf (x) − f (x) = −23/16 , g (4) = 2 x 16

F (x) = 5f (x) + 2g (x), F (2) = 5(4) + 2(−5) = 10 F (x) = f (x) − 3g (x), F (2) = 4 − 3(−5) = 19 F (x) = f (x)g (x) + g(x)f (x), F (2) = (−1)(−5) + (1)(4) = 9 F (x) = [g(x)f (x) − f (x)g (x)]/g 2 (x), F (2) = [(1)(4) − (−1)(−5)]/(1)2 = −1

41. (a) (b) (c) (d)

2f (x) − (2x + 1)f (x) 2(−2) − 7(4) , g (3) = = −8 f 2 (x) (−2)2

(b) g (x) =

40. (a) g (x) = 6x − 5f (x), g (3) = 6(3) − 5(4) = −2

38.

(b)

dV = 4π(5)2 = 100π dr r=5

us uf

37. (a)

dV = 4πr2 dr

Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Exercise Set 3.3

(4 + g(x))f (x) − f (x)g (x) (4 − 3)(−1) − 10(2) = = −21 (4 + g(x))2 (4 − 3)2

(d) F (x) =

42. (a) F (x) = 6f (x) − 5g (x), F (π) = 6(−1) − 5(2) = −16 (b) F (x) = f (x) + g(x) + x(f (x) + g (x)), F (π) = 10 − 3 + π(−1 + 2) = 7 + π (c) F (x) = 2f (x)g (x) + 2f (x)g(x) = 2(20) + 2(3) = 46

43. y − 2 = 5(x + 3), y = 5x + 17

dy/dx = 21x2 − 10x + 1, d2 y/dx2 = 42x − 10 dy/dx = 24x − 2, d2 y/dx2 = 24 dy/dx = −1/x2 , d2 y/dx2 = 2/x3 y = 35x5 − 16x3 − 3x, dy/dx = 175x4 − 48x2 − 3, d2 y/dx2 = 700x3 − 96x

45. (a) (b) (c) (d)

2 dy 2 dy (1 + x)(−1) − (1 − x)(1) 1 =− , = − and y = − for x = 2 so an equation 44. = 2 2 dx (1 + x) (1 + x) dx x=2 9 3 2 1 2 1 = − (x − 2), or y = − x + . of the tangent line is y − − 3 9 9 9

uh a

46. (a) y = 28x6 − 15x2 + 2, y = 168x5 − 30x (b) y = 3, y = 0 2 4 (c) y = 2 , y = − 3 5x 5x (d) y = 2x4 + 3x3 − 10x − 15, y = 8x3 + 9x2 − 10, y = 24x2 + 18x

47. (a) y = −5x−6 + 5x4 , y = 30x−7 + 20x3 , y = −210x−8 + 60x2 (b) y = x−1 , y = −x−2 , y = 2x−3 , y = −6x−4 (c) y = 3ax2 + b, y = 6ax, y = 6a

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Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Chapter 3

us uf

48. (a) dy/dx = 10x − 4, d2 y/dx2 = 10, d3 y/dx3 = 0 (b) dy/dx = −6x−3 − 4x−2 + 1, d2 y/dx2 = 18x−4 + 8x−3 , d3 y/dx3 = −72x−5 − 24x−4 (c) dy/dx = 4ax3 + 2bx, d2 y/dx2 = 12ax2 + 2b, d3 y/dx3 = 24ax 49. (a) f (x) = 6x, f (x) = 6, f (x) = 0, f (2) = 0 dy d2 y d2 y 4 3 = 30x − 8x, (b) = 120x − 8, = 112 dx2 dx2 x=1 dx

d −3

d4 −3

d2 −3

d3 −3

−5 −6 x = 12x x = −60x x = 360x−7 , , = −3x−4 , x , dx4 dx2 dx3 dx d4 −3 = 360 x dx4 x=1

(c)

50. (a) y = 16x3 + 6x2 , y = 48x2 + 12x, y = 96x + 12, y (0) = 12

dy d2 y d3 y d4 y (b) y = 6x−4 , = 120x−6 , = −720x−7 , = 5040x−8 , = −24x−5 , 2 3 dx dx dx dx4 d4 y = 5040 dx4 x=1 51. y = 3x2 + 3, y = 6x, and y = 6 so y + xy − 2y = 6 + x(6x) − 2(3x2 + 3) = 6 + 6x2 − 6x2 − 6 = 0

52. y = x−1 , y = −x−2 , y = 2x−3 so x3 y + x2 y − xy = x3 (2x−3 ) + x2 (−x−2 ) − x(x−1 ) = 2 − 1 − 1 = 0 53. F (x) = xf (x) + f (x), F (x) = xf (x) + f (x) + f (x) = xf (x) + 2f (x)

54. (a) F (x) = xf (x) + 3f (x) (b) Assume that F n) (x) = xf (n) (x) + nf (n−1) (x) for some n (for instance n = 3, as in part (a)). Then F (n+1) (x) = xf (n+1) (x) + (1 + n)f (n) (x) = xf (n+1) (x) + (n + 1)f (n) (x), which is an inductive proof. dy = 0, 55. The graph has a horizontal tangent at points where dx dy but = x2 − 3x + 2 = (x − 1)(x − 2) = 0 if x = 1, 2. The dx corresponding values of y are 5/6 and 2/3 so the tangent line is horizontal at (1, 5/6) and (2, 2/3).

1.5

3 0

9 − x2 dy dy = 2 = 0 when x2 = 9 so x = ±3. ; 2 dx (x + 9) dx The points are (3, 1/6) and (−3, −1/6).

0.2

-5.5

5.5

-0.2

uh a

56.

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Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Exercise Set 3.3

57. The y-intercept is −2 so the point (0, −2) is on the graph; −2 = a(0)2 + b(0) + c, c = −2. The x-intercept is 1 so the point (1,0) is on the graph; 0 = a + b − 2. The slope is dy/dx = 2ax + b; at x = 0 the slope is b so b = −1, thus a = 3. The function is y = 3x2 − x − 2.

us uf

dy = 2x 58. Let P (x0 , y0 ) be the point where y = x2 +k is tangent to y = 2x. The slope of the curve is dx and the slope of the line is 2 thus at P , 2x0 = 2 so x0 = 1. But P is on the line, so y0 = 2x0 = 2. Because P is also on the curve we get y0 = x20 + k so k = y0 − x20 = 2 − (1)2 = 1.

59. The points (−1, 1) and (2, 4) are on the secant line so its slope is (4 − 1)/(2 + 1) = 1. The slope of the tangent line to y = x2 is y = 2x so 2x = 1, x = 1/2.

60. The points √ (1, 1) and (4, √ 2) are on the √ secant line √ so its slope is 1/3. The slope of the tangent line to y = x is y = 1/(2 x) so 1/(2 x) = 1/3, 2 x = 3, x = 9/4.

61. y = −2x, so at any point (x0 , y0 ) on y = 1 − x2 the tangent line is y − y0 = −2x0 (x − x0 ), or 2 2 y = −2x0 x + x20 + 1. The point (2, 0) is to be √ on the line, so 0 = −4x0 + x0 + 1, x0 − 4x0 + 1 = 0. √ 4 ± 16 − 4 = 2 ± 3. Use the quadratic formula to get x0 = 2

62. Let P1 (x1 , ax21 ) and P2 (x2 , ax22 ) be the points of tangency. y = 2ax so the tangent lines at P1 and P2 are y − ax21 = 2ax1 (x − x1 ) and y − ax22 = 2ax2 (x − x2 ). Solve for x to get x = 12 (x1 + x2 ) which is the x-coordinate of a point on the vertical line halfway between P1 and P2 .

63. y = 3ax2 + b; the tangent line at x = x0 is y − y0 = (3ax20 + b)(x − x0 ) where y0 = ax30 + bx0 . Solve with y = ax3 + bx to get

(ax3 + bx) − (ax30 + bx0 ) = (3ax20 + b)(x − x0 ) ax3 + bx − ax30 − bx0 = 3ax20 x − 3ax30 + bx − bx0 x3 − 3x20 x + 2x30 = 0 (x − x0 )(x2 + xx0 − 2x20 ) = 0 (x − x0 )2 (x + 2x0 ) = 0, so x = −2x0 . 64. Let (x0 , y0 ) be the point of tangency. Refer to the solution to Exercise 65 to see that the endpoints of the line segment are at (2x0 , 0) and (0, 2y0 ), so (x0 , y0 ) is the midpoint of the segment. 1 1 x 2 ; the tangent line at x = x0 is y − y0 = − 2 (x − x0 ), or y = − 2 + . The tangent line 2 x x0 x0 x0 1 crosses the x-axis at 2x0 , the y-axis at 2/x0 , so that the area of the triangle is (2/x0 )(2x0 ) = 2. 2

65. y = −

66. f (x) = 3ax2 + 2bx + c; there is a horizontal √ tangent where f (x) = 0. Use the quadratic formula 2 2 on 3ax + 2bx + c = 0 to get x = (−b ± b − 3ac)/(3a) which gives two real solutions, one real solution, or none if

(a) b2 − 3ac > 0

uh a

67. F = GmM r−2 ,

(b) b2 − 3ac = 0

dF 2GmM = −2GmM r−3 = − dr r3

68. dR/dT = 0.04124 − 3.558 × 10−5 T which decreases as T increases from 0 to 700. When T = 0, dR/dT = 0.04124 Ω/◦ C; when T = 700, dR/dT = 0.01633 Ω/◦ C. The resistance is most sensitive to temperature changes at T = 0◦ C, least sensitive at T = 700◦ C.

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Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Chapter 3

x2 − 4 ; (x2 + 4)2 f (x) > 0 when x2 < 4, i.e. on −2 < x < 2

70. f (x) = −5

1.5 -6

6 -5

-1.5

-6

69. f (x) = 1 + 1/x2 > 0 for all x = 0

us uf

71. (f · g · h) = [(f · g) · h] = (f · g)h + h(f · g) = (f · g)h + h[f g + f g] = f gh + f g h + f gh

72. (f1 f2 · · · fn ) = (f1 f2 · · · fn ) + (f1 f2 · · · fn ) + · · · + (f1 f2 · · · fn ) 73. (a) 2(1 + x−1 )(x−3 + 7) + (2x + 1)(−x−2 )(x−3 + 7) + (2x + 1)(1 + x−1 )(−3x−4 )

(b) (x7 + 2x − 3)3 = (x7 + 2x − 3)(x7 + 2x − 3)(x7 + 2x − 3) so d 7 (x + 2x − 3)3 = (7x6 + 2)(x7 + 2x − 3)(x7 + 2x − 3) dx +(x7 + 2x − 3)(7x6 + 2)(x7 + 2x − 3)

+(x7 + 2x − 3)(x7 + 2x − 3)(7x6 + 2)

= 3(7x6 + 2)(x7 + 2x − 3)2

74. (a) −5x−6 (x2 + 2x)(4 − 3x)(2x9 + 1) + x−5 (2x + 2)(4 − 3x)(2x9 + 1) +x−5 (x2 + 2x)(−3)(2x9 + 1) + x−5 (x2 + 2x)(4 − 3x)(18x8 ) (b) (x2 + 1)50 = (x2 + 1)(x2 + 1) · · · (x2 + 1), where (x2 + 1) occurs 50 times so

d 2 (x + 1)50 = [(2x)(x2 + 1) · · · (x2 + 1)] + [(x2 + 1)(2x) · · · (x2 + 1)] dx + · · · + [(x2 + 1)(x2 + 1) · · · (2x)] = 2x(x2 + 1)49 + 2x(x2 + 1)49 + · · · + 2x(x2 + 1)49

= 100x(x2 + 1)49 because 2x(x2 + 1)49 occurs 50 times. 75. f is continuous at 1 because lim f (x) = lim f (x) = f (1); also lim f (x) = lim (2x + 1) = 3 − + − − x→1

x→1

and lim+ f (x) = lim+ 3 = 3 so f is diﬀerentiable at 1. x→1

x→1

76. f is not continuous at x = 9 because lim f (x) = −63 and lim f (x) = 36. − + x→9

x→9

f cannot be diﬀerentiable at x = 9, for if it were, then f would also be continuous, which it is not. 77. f is continuous at 1 because lim f (x) = lim f (x) = f (1), also lim f (x) = lim 2x = 2 and − + − − x→1

x→1

uh a

1 1 lim f (x) = lim+ √ = so f is not diﬀerentiable at 1. 2 x→1 2 x x→1+

78. f is continuous at 1/2 because

lim f (x) =

x→1/2−

lim f (x) =

x→1/2−

lim 3x2 = 3/4 and

x→1/2−

x→1

lim f (x) = f (1/2), also

x→1/2+

lim f (x) =

x→1/2+

x→1

lim 3x/2 = 3/4 so f (1/2) = 3/4, and f

x→1/2+

is diﬀerentiable at x = 1/2.

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Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Exercise Set 3.4

79. (a) f (x) = 3x − 2 if x ≥ 2/3, f (x) = −3x + 2 if x < 2/3 so f is diﬀerentiable everywhere except perhaps at 2/3. f is continuous at 2/3, also lim f (x) = lim (−3) = −3 and x→2/3−

lim (3) = 3 so f is not diﬀerentiable at x = 2/3.

x→2/3+

us uf

lim f (x) =

x→2/3+

x→2/3−

(b) f (x) = x2 − 4 if |x| ≥ 2, f (x) = −x2 + 4 if |x| < 2 so f is diﬀerentiable everywhere except perhaps at ±2. f is continuous at −2 and 2, also lim f (x) = lim (−2x) = −4 x→2−

x→2−

and lim+ f (x) = lim+ (2x) = 4 so f is not diﬀerentiable at x = 2. Similarly, f is not x→2

x→2

80. (a) f (x) = −(1)x−2 , f (x) = (2 · 1)x−3 , f (x) = −(3 · 2 · 1)x−4 f (n) (x) = (−1)n

n(n − 1)(n − 2) · · · 1 xn+1

(b) f (x) = −2x−3 , f (x) = (3 · 2)x−4 , f (x) = −(4 · 3 · 2)x−5

(n + 1)(n)(n − 1) · · · 2 xn+2 d2 d d d d d d2 d [f (x)] = c [cf (x)] = c [f (x)] = c [f (x)] [cf (x)] = dx2 dx dx dx dx dx2 dx dx d2 d d d d d [f (x) + g(x)] = [f (x)] + [g(x)] [f (x) + g(x)] = dx dx dx dx dx dx2

f (n) (x) = (−1)n

d2 d2 [f (x)] + 2 [g(x)] 2 dx dx

81. (a)

diﬀerentiable at x = −2.

(b) yes, by repeated application of the procedure illustrated in Part (a) 82. (f · g) = f g + gf , (f · g) = f g + g f + gf + f g = f g + 2f g + f g

83. (a) f (x) = nxn−1 , f (x) = n(n − 1)xn−2 , f (x) = n(n − 1)(n − 2)xn−3 , . . ., f (n) (x) = n(n − 1)(n − 2) · · · 1 (b) from Part (a), f (k) (x) = k(k − 1)(k − 2) · · · 1 so f (k+1) (x) = 0 thus f (n) (x) = 0 if n > k (c) from Parts (a) and (b), f (n) (x) = an n(n − 1)(n − 2) · · · 1 f (2 + h) − f (2) = f (2); f (x) = 8x7 − 2, f (x) = 56x6 , so f (2) = 56(26 ) = 3584. h→0 h

84.

lim

85. (a) If a function is diﬀerentiable at a point then it is continuous at that point, thus f is continuous on (a, b) and consequently so is f . (b) f and all its derivatives up to f (n−1) (x) are continuous on (a, b)

uh a

EXERCISE SET 3.4 1. f (x) = −2 sin x − 3 cos x

2. f (x) = sin x(− sin x) + cos x(cos x) = cos2 x − sin2 x = cos 2x 3. f (x) =

x(cos x) − (sin x)(1) x cos x − sin x = 2 x x2

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Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Chapter 3

4. f (x) = x2 (− sin x) + (cos x)(2x) = −x2 sin x + 2x cos x

us uf

cos x cot x x(− csc2 x) − (cot x)(1) x csc2 x + cot x (because = cot x), f (x) = =− 2 x sin x x x2 √ 7. f (x) = sec x tan x − 2 sec2 x 6. f (x) =

(1 + tan x)(sec x tan x) − (sec x)(sec2 x) sec x tan x + sec x tan2 x − sec3 x = 2 (1 + tan x) (1 + tan x)2 sec x(tan x − 1) sec x(tan x + tan2 x − sec2 x) = 2 (1 + tan x)2 (1 + tan x)

10. f (x) =

8. f (x) = (x2 + 1) sec x tan x + (sec x)(2x) = (x2 + 1) sec x tan x + 2x sec x 9. f (x) = sec x(sec2 x) + (tan x)(sec x tan x) = sec3 x + sec x tan2 x

11. f (x) = (csc x)(− csc2 x) + (cot x)(− csc x cot x) = − csc3 x − csc x cot2 x 12. f (x) = 1 + 4 csc x cot x − 2 csc2 x 13. f (x) =

(1 + csc x)(− csc2 x) − cot x(0 − csc x cot x) csc x(− csc x − csc2 x + cot2 x) = but (1 + csc x)2 (1 + csc x)2

csc x(1 + sec2 x) tan x(− csc x cot x) − csc x(sec2 x) =− 2 tan x tan2 x

14. f (x) =

csc x(− csc x − 1) csc x =− 2 (1 + csc x) 1 + csc x

1 + cot2 x = csc2 x (identity) thus cot2 x − csc2 x = −1 so f (x) =

15. f (x) = sin2 x + cos2 x = 1 (identity) so f (x) = 0 1 = tan x, so f (x) = sec2 x cot x

17.

f (x) =

tan x (because sin x sec x = (sin x)(1/ cos x) = tan x), 1 + x tan x

f (x) =

(1 + x tan x)(sec2 x) − tan x[x(sec2 x) + (tan x)(1)] (1 + x tan x)2

f (x) =

(x2 + 1) cot x (because cos x csc x = (cos x)(1/ sin x) = cot x), 3 − cot x

uh a

18.

sec2 x − tan2 x 1 = (because sec2 x − tan2 x = 1) 2 (1 + x tan x) (1 + x tan x)2

16. f (x) =

f (x) =

5. f (x) = x3 (cos x) + (sin x)(3x2 ) − 5(− sin x) = x3 cos x + (3x2 + 5) sin x

(3 − cot x)[2x cot x − (x2 + 1) csc2 x] − (x2 + 1) cot x csc2 x (3 − cot x)2 6x cot x − 2x cot2 x − 3(x2 + 1) csc2 x (3 − cot x)2

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Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Exercise Set 3.4

19. dy/dx = −x sin x + cos x, d2 y/dx2 = −x cos x − sin x − sin x = −x cos x − 2 sin x

20. dy/dx = − csc x cot x, d2 y/dx2 = −[(csc x)(− csc2 x) + (cot x)(− csc x cot x)] = csc3 x + csc x cot2 x

us uf

21. dy/dx = x(cos x) + (sin x)(1) − 3(− sin x) = x cos x + 4 sin x, d2 y/dx2 = x(− sin x) + (cos x)(1) + 4 cos x = −x sin x + 5 cos x

22. dy/dx = x2 (− sin x) + (cos x)(2x) + 4 cos x = −x2 sin x + 2x cos x + 4 cos x, d2 y/dx2 = −[x2 (cos x) + (sin x)(2x)] + 2[x(− sin x) + cos x] − 4 sin x = (2 − x2 ) cos x − 4(x + 1) sin x

23. dy/dx = (sin x)(− sin x) + (cos x)(cos x) = cos2 x − sin2 x, d2 y/dx2 = (cos x)(− sin x) + (cos x)(− sin x) − [(sin x)(cos x) + (sin x)(cos x)] = −4 sin x cos x 24. dy/dx = sec2 x; d2 y/dx2 = 2 sec2 x tan x

25. Let f (x) = tan x, then f (x) = sec2 x. (a) f (0) = 0 and f (0) = 1 so y − 0 = (1)(x − 0), y = x. π

π = 1 and f = 2 so y − 1 = 2 x − , y = 2x − + 1. (b) f 4 4 4 2 π

π = −1 and f − = 2 so y + 1 = 2 x + , y = 2x + − 1. (c) f − 4 4 4 2

26. Let f (x) = sin x, then f (x) = cos x. (a) f (0) = 0 and f (0) = 1 so y − 0 = (1)(x − 0), y = x (b) f (π) = 0 and f (π) = −1 so y − 0 = (−1)(x − π), y = −x + π π

1 1 1 1 1 π

π 1 (c) f = √ and f = √ so y − √ = √ x − , y = √ x− √ + √ 4 4 4 2 2 2 2 2 2 4 2

27. (a) If y = x sin x then y = sin x + x cos x and y = 2 cos x − x sin x so y + y = 2 cos x. (b) If y = x sin x then y = sin x+x cos x and y = 2 cos x−x sin x so y +y = 2 cos x; diﬀerentiate twice more to get y (4) + y = −2 cos x. 28. (a) If y = cos x then y = − sin x and y = − cos x so y + y = (− cos x) + (cos x) = 0; if y = sin x then y = cos x and y = − sin x so y + y = (− sin x) + (sin x) = 0. (b) y = A cos x − B sin x, y = −A sin x − B cos x so y + y = (−A sin x − B cos x) + (A sin x + B cos x) = 0.

f (x) = cos x = 0 at x = ±π/2, ±3π/2. f (x) = 1 − sin x = 0 at x = −3π/2, π/2. f (x) = sec2 x ≥ 1 always, so no horizontal tangent line. f (x) = sec x tan x = 0 when sin x = 0, x = ±2π, ±π, 0

30. (a)

0.5

29. (a) (b) (c) (d)

uh a

-0.5

(b) y = sin x cos x = (1/2) sin 2x and y = cos 2x. So y = 0 when 2x = (2n + 1)π/2 for n = 0, 1, 2, 3 or x = π/4, 3π/4, 5π/4, 7π/4

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Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Chapter 3

33. D = 50 tan Î¸,â&#x2C6;&#x161; dD/dÎ¸ = 50 sec2 Î¸; if Î¸ = 45â&#x2014;Ś , then dD/dÎ¸ = 50( 2)2 = 100 m/rad = 5Ď&#x20AC;/9 m/deg â&#x2030;&#x2C6; 1.75 m/deg

us uf

â&#x2014;Ś 32. s = 3800 csc Î¸, ds/dÎ¸ â&#x2C6;&#x161;= â&#x2C6;&#x2019;3800 cscâ&#x2C6;&#x161;Î¸ cot Î¸; if Î¸ = 30 ,â&#x2C6;&#x161;then ds/dÎ¸ = â&#x2C6;&#x2019;3800(2)( 3) = â&#x2C6;&#x2019;7600 3 ft/rad = â&#x2C6;&#x2019;380 3Ď&#x20AC;/9 ft/deg â&#x2030;&#x2C6; â&#x2C6;&#x2019;230 ft/deg

31. x = 10 sin Î¸, dx/dÎ¸ = 10 cos Î¸; if Î¸ = 60â&#x2014;Ś , then dx/dÎ¸ = 10(1/2) = 5 ft/rad = Ď&#x20AC;/36 ft/deg â&#x2030;&#x2C6; 0.087 ft/deg

(b)

d100 d4k cos x = 4k cos x = cos x 100 dx dx

d [x sin x] = x cos x + sin x dx

d2 [x sin x] = â&#x2C6;&#x2019;x sin x + 2 cos x dx2

d3 [x sin x] = â&#x2C6;&#x2019;x cos x â&#x2C6;&#x2019; 3 sin x dx3

d4 [x sin x] = x sin x â&#x2C6;&#x2019; 4 cos x dx4

36.

d3 d4Âˇ21 d3 d4k d87 d4 sin x = â&#x2C6;&#x2019; cos x sin x = sin x = sin x = sin x, so sin x = sin x; dx3 dx4k dx87 dx3 dx4Âˇ21 dx4

35. (a)

34. (a) From the right triangle shown, sin Î¸ = r/(r + h) so r + h = r csc Î¸, h = r(csc Î¸ â&#x2C6;&#x2019; 1). (b) dh/dÎ¸ = â&#x2C6;&#x2019;r csc Î¸ cotâ&#x2C6;&#x161;Î¸; if Î¸ = 30â&#x2014;Ś , then dh/dÎ¸ = â&#x2C6;&#x2019;6378(2)( 3) â&#x2030;&#x2C6; â&#x2C6;&#x2019;22, 094 km/rad â&#x2030;&#x2C6; â&#x2C6;&#x2019;386 km/deg

By mathematical induction one can show

d17 [x sin x] = x cos x + 17 sin x dx17

all x x = Ď&#x20AC;/2 + nĎ&#x20AC;, n = 0, Âą1, Âą2, . . . x = Ď&#x20AC;/2 + nĎ&#x20AC;, n = 0, Âą1, Âą2, . . . x = (2n + 1)Ď&#x20AC;, n = 0, Âą1, Âą2, . . . all x

(b) (d) (f ) (h)

all x x = nĎ&#x20AC;, n = 0, Âą1, Âą2, . . . x = nĎ&#x20AC;, n = 0, Âą1, Âą2, . . . x = nĎ&#x20AC;/2, n = 0, Âą1, Âą2, . . .

37. (a) (c) (e) (g) (i)

d4k+3 [x sin x] = â&#x2C6;&#x2019;x cos x â&#x2C6;&#x2019; (4k + 3) sin x; dx4k+3

d4k+2 [x sin x] = â&#x2C6;&#x2019;x sin x + (4k + 2) cos x; dx4k+2 Since 17 = 4 Âˇ 4 + 1,

d4k+1 [x sin x] = x cos x + (4k + 1) sin x; dx4k+1

d4k [x sin x] = x sin x â&#x2C6;&#x2019; (4k) cos x; dx4k

cos(x + h) â&#x2C6;&#x2019; cos x cos x cos h â&#x2C6;&#x2019; sin x sin h â&#x2C6;&#x2019; cos x d [cos x] = lim = lim hâ&#x2020;&#x2019;0 hâ&#x2020;&#x2019;0 dx h h sin h cos h â&#x2C6;&#x2019; 1 â&#x2C6;&#x2019; sin x = (cos x)(0) â&#x2C6;&#x2019; (sin x)(1) = â&#x2C6;&#x2019; sin x = lim cos x hâ&#x2020;&#x2019;0 h h 1 cos x(0) â&#x2C6;&#x2019; (1)(â&#x2C6;&#x2019; sin x) d sin x d = (b) = = sec x tan x [sec x] = dx dx cos x cos2 x cos2 x

38. (a)

d cos x sin x(â&#x2C6;&#x2019; sin x) â&#x2C6;&#x2019; cos x(cos x) d [cot x] = = dx dx sin x sin2 x

uh a M

(c)

â&#x2C6;&#x2019; sin2 x â&#x2C6;&#x2019; cos2 x â&#x2C6;&#x2019;1 = = â&#x2C6;&#x2019; csc2 x 2 sin2 x sin x 1 (sin x)(0) â&#x2C6;&#x2019; (1)(cos x) d cos x d = (d) = â&#x2C6;&#x2019; 2 = â&#x2C6;&#x2019; csc x cot x [csc x] = 2 dx dx sin x sin x sin x =

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Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Exercise Set 3.5

39. f (x) = − sin x, f (x) = − cos x, f (x) = sin x, and f (4) (x) = cos x with higher order derivatives repeating this pattern, so f (n) (x) = sin x for n = 3, 7, 11, . . .

40. (a)

lim

h→0

tan h = lim h→0 h

= lim

h→0

sin h h cos h

sin h cos h h

us uf

1 =1 1

tan x + tan h − tan x tan(x + h) − tan x d 1 = lim − tan x tan h (b) [tan x] = lim h→0 h→0 h h dx

tan h tan h sec2 x h = sec2 x lim = lim h→0 1 − tan x tan h h→0 h(1 − tan x tan h) tan h h→0 h = sec2 x = sec2 x lim (1 − tan x tan h)

lim

x→0

h→0

41.

tan x + tan h − tan x + tan2 x tan h tan h(1 + tan2 x) = lim h→0 h(1 − tan x tan h) h→0 h(1 − tan x tan h)

= lim

tan(x + y) − tan y tan(y + h) − tan y d = lim = (tan y) = sec2 y h→0 x h dy

180 t and cos h = cos t, sin h = sin t. π cos h − 1 cos t − 1 cos t − 1 π (a) lim = lim = lim =0 t→0 180t/π h→0 h 180 t→0 t

43. Let t be the radian measure, then h =

π π sin h sin t sin t = lim = lim = t→0 180t/π h→0 h 180 t→0 t 180

(c)

π cos h − 1 sin h d + cos x lim = (sin x)(0) + (cos x)(π/180) = cos x [sin x] = sin x lim h→0 h→0 h h 180 dx

(b)

EXERCISE SET 3.5

lim

1. (f ◦ g) (x) = f (g(x))g (x) so (f ◦ g) (0) = f (g(0))g (0) = f (0)(3) = (2)(3) = 6

2. (f ◦ g) (2) = f (g(2))g (2) = 5(−3) = −15

3. (a) (f ◦ g)(x) = f (g(x)) = (2x − 3)5 and (f ◦ g) (x) = f (g(x)g (x) = 5(2x − 3)4 (2) = 10(2x − 3)4 (b) (g ◦ f )(x) = g(f (x)) = 2x5 − 3 and (g ◦ f ) (x) = g (f (x))f (x) = 2(5x4 ) = 10x4

uh a

√ 5 4. (a) (f ◦ g)(x) = 5 4 + cos x and (f ◦ g) (x) = f (g(x))g (x) = √ (− sin x) 2 4 + cos x √ √ 5 (b) (g ◦ f )(x) = 4 + cos(5 x) and (g ◦ f ) (x) = g (f (x))f (x) = − sin(5 x) √ 2 x

5. (a) F (x) = f (g(x))g (x) = f (g(3))g (3) = −1(7) = −7 (b) G (x) = g (f (x))f (x) = g (f (3))f (3) = 4(−2) = −8

6. (a) F (x) = f (g(x))g (x), F (−1) = f (g(−1))g (−1) = f (2)(−3) = (4)(−3) = −12 (b) G (x) = g (f (x))f (x), G (−1) = g (f (−1))f (−1) = −5(3) = −15

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Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Chapter 3

7. f (x) = 37(x3 + 2x)36

d 3 (x + 2x) = 37(x3 + 2x)36 (3x2 + 2) dx

d (3x2 + 2x â&#x2C6;&#x2019; 1) = 6(3x2 + 2x â&#x2C6;&#x2019; 1)5 (6x + 2) = 12(3x2 + 2x â&#x2C6;&#x2019; 1)5 (3x + 1) dx â&#x2C6;&#x2019;3 â&#x2C6;&#x2019;3 d 7 7 7 7 3 3 3 2 x â&#x2C6;&#x2019; = â&#x2C6;&#x2019;2 x â&#x2C6;&#x2019; 3x + 2 9. f (x) = â&#x2C6;&#x2019;2 x â&#x2C6;&#x2019; x dx x x x d 5 9(5x4 â&#x2C6;&#x2019; 1) (x â&#x2C6;&#x2019; x + 1) = â&#x2C6;&#x2019;9(x5 â&#x2C6;&#x2019; x + 1)â&#x2C6;&#x2019;10 (5x4 â&#x2C6;&#x2019; 1) = â&#x2C6;&#x2019; 5 dx (x â&#x2C6;&#x2019; x + 1)10

11. f (x) = 4(3x2 â&#x2C6;&#x2019; 2x + 1)â&#x2C6;&#x2019;3 , f (x) = â&#x2C6;&#x2019;12(3x2 â&#x2C6;&#x2019; 2x + 1)â&#x2C6;&#x2019;4

f (x) = â&#x2C6;&#x2019;9(x5 â&#x2C6;&#x2019; x + 1)â&#x2C6;&#x2019;10

d 24(1 â&#x2C6;&#x2019; 3x) (3x2 â&#x2C6;&#x2019; 2x + 1) = â&#x2C6;&#x2019;12(3x2 â&#x2C6;&#x2019; 2x + 1)â&#x2C6;&#x2019;4 (6x â&#x2C6;&#x2019; 2) = (3x2 â&#x2C6;&#x2019; 2x + 1)4 dx

â&#x2C6;&#x161; d 1 3 13. f (x) = (4 + 3 x) = â&#x2C6;&#x161; â&#x2C6;&#x161; â&#x2C6;&#x161; dx 2 4+3 x 4 x 4+3 x d (sin x) = 3 sin2 x cos x dx

15. f (x) = cos(x3 )

d 3 (x ) = 3x2 cos(x3 ) dx

14. f (x) = 3 sin2 x

1 3x2 â&#x2C6;&#x2019; 2 d 3 12. f (x) = â&#x2C6;&#x161; (x â&#x2C6;&#x2019; 2x + 5) = â&#x2C6;&#x161; 2 x3 â&#x2C6;&#x2019; 2x + 5 dx 2 x3 â&#x2C6;&#x2019; 2x + 5

10. f (x) = (x5 â&#x2C6;&#x2019; x + 1)â&#x2C6;&#x2019;9 ,

us uf

8. f (x) = 6(3x2 + 2x â&#x2C6;&#x2019; 1)5

â&#x2C6;&#x161; â&#x2C6;&#x161; â&#x2C6;&#x161; â&#x2C6;&#x161; â&#x2C6;&#x161; d â&#x2C6;&#x161; d 3 cos(3 x) sin(3 x) â&#x2C6;&#x161; 16. f (x) = 2 cos(3 x) [cos(3 x)] = â&#x2C6;&#x2019;2 cos(3 x) sin(3 x) (3 x) = â&#x2C6;&#x2019; dx dx x d (cos x) = 20 cos4 x(â&#x2C6;&#x2019; sin x) = â&#x2C6;&#x2019;20 cos4 x sin x dx

18. f (x) = â&#x2C6;&#x2019; csc(x3 ) cot(x3 )

d 3 (x ) = â&#x2C6;&#x2019;3x2 csc(x3 ) cot(x3 ) dx

d 2 (1/x2 ) = â&#x2C6;&#x2019; 3 cos(1/x2 ) dx x

19. f (x) = cos(1/x2 )

17. f (x) = 20 cos4 x

â&#x2C6;&#x161;

20. f (x) = 4 tan3 (x3 )

d d [tan(x3 )] = 4 tan3 (x3 ) sec2 (x3 ) (x3 ) = 12x2 tan3 (x3 ) sec2 (x3 ) dx dx

d d [sec(x7 )] = 4 sec(x7 ) sec(x7 ) tan(x7 ) (x7 ) = 28x6 sec2 (x7 ) tan(x7 ) dx dx d x (x + 1)(1) â&#x2C6;&#x2019; x(1) x x x 2 2 = 3 cos â&#x2C6;&#x2019; sin 22. f (x) = 3 cos cos x + 1 dx x+1 x+1 x+1 (x + 1)2 x x 3 sin =â&#x2C6;&#x2019; cos2 2 (x + 1) x+1 x+1

uh a

21. f (x) = 4 sec(x7 )

5 sin(5x) 1 d [cos(5x)] = â&#x2C6;&#x2019; 23. f (x) = 2 cos(5x) dx 2 cos(5x)

d 1 3 â&#x2C6;&#x2019; 8 sin(4x) cos(4x) 24. f (x) = [3x â&#x2C6;&#x2019; sin2 (4x)] = 2 3x â&#x2C6;&#x2019; sin2 (4x) dx 2 3x â&#x2C6;&#x2019; sin2 (4x)

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Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Exercise Set 3.5

−5 d 4 x − sec(4x2 − 2) 26. f (x) = −4 x4 − sec(4x2 − 2) dx 4

d −5 = −4 x − sec(4x2 − 2) 4x3 − sec(4x2 − 2) tan(4x2 − 2) (4x2 − 2) dx 4

−5 = −16x x − sec(4x2 − 2) x2 − 2 sec(4x2 − 2) tan(4x2 − 2)

us uf

−4 d

x + csc(x3 + 3) 25. f (x) = −3 x + csc(x3 + 3) dx

d −4 = −3 x + csc(x3 + 3) 1 − csc(x3 + 3) cot(x3 + 3) (x3 + 3) dx

−4 1 − 3x2 csc(x3 + 3) cot(x3 + 3) = −3 x + csc(x3 + 3)

dy d = x3 (2 sin 5x) (sin 5x) + 3x2 sin2 5x = 10x3 sin 5x cos 5x + 3x2 sin2 5x dx dx

28.

√ √ √ √ √ √ √ 1 dy 1 3 1 = x 3 tan2 ( x) sec2 ( x) √ + √ tan3 ( x) = tan2 ( x) sec2 ( x) + √ tan3 ( x) 2 dx 2 x 2 x 2 x

29.

dy = x5 sec dx

30.

sec(3x + 1) cos x − 3 sin x sec(3x + 1) tan(3x + 1) dy = = cos x cos(3x + 1) − 3 sin x sin(3x + 1) dx sec2 (3x + 1)

31.

d dy = − sin(cos x) (cos x) = − sin(cos x)(− sin x) = sin(cos x) sin x dx dx

32.

dy d = cos(tan 3x) (tan 3x) = 3 sec2 3x cos(tan 3x) dx dx

33.

1 d 1 1 1 tan + sec (5x4 ) x x dx x x 1 1 1 1 tan − 2 + 5x4 sec = x5 sec x x x x 1 1 1 3 4 = −x sec tan + 5x sec x x x

27.

dy d d = 3 cos2 (sin 2x) [cos(sin 2x)] = 3 cos2 (sin 2x)[− sin(sin 2x)] (sin 2x) dx dx dx

= −6 cos2 (sin 2x) sin(sin 2x) cos 2x 2 2 dy (1 − cot x2 )(−2x csc x2 cot x2 ) − (1 + csc x2 )(2x csc2 x2 ) 2 1 + cot x + csc x = −2x csc x = dx (1 − cot x2 )2 (1 − cot x2 )2

35.

d d dy = (5x + 8)13 12(x3 + 7x)11 (x3 + 7x) + (x3 + 7x)12 13(5x + 8)12 (5x + 8) dx dx dx

uh a

34.

= 12(5x + 8)13 (x3 + 7x)11 (3x2 + 7) + 65(x3 + 7x)12 (5x + 8)12

36.

dy = (2x − 5)2 3(x2 + 4)2 (2x) + (x2 + 4)3 2(2x − 5)(2) dx = 6x(2x − 5)2 (x2 + 4)2 + 4(2x − 5)(x2 + 4)3 = 2(2x − 5)(x2 + 4)2 (8x2 − 15x + 8)

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37.

2 2 x−5 x−5 d x−5 11 33(x − 5)2 dy =3 =3 · = 2x + 1 dx 2x + 1 2x + 1 (2x + 1)2 (2x + 1)4 dx

38.

dy = 17 dx

= 17

d dx

1 + x2 1 − x2

(1 − x2 )(2x) − (1 + x2 )(−2x) (1 − x2 )2

2(2x + 3)2 (4x2 − 1)7 [3(4x2 − 1) − 32x(2x + 3)] 2(2x + 3)2 (52x2 + 96x + 3) = − (4x2 − 1)16 (4x2 − 1)9

d dy = 12[1 + sin3 (x5 )]11 [1 + sin3 (x5 )] dx dx

d sin(x5 ) = 180x4 [1 + sin3 (x5 )]11 sin2 (x5 ) cos(x5 ) dx

dy = 4 tan3 dx

4 d

dy x sin 2x tan4 (x7 ) = 5 x sin 2x + tan4 (x7 ) dx dx

4 d d = 5 x sin 2x + tan4 (x7 ) x cos 2x (2x) + sin 2x + 4 tan3 (x7 ) tan(x7 ) dx dx

4 = 5 x sin 2x + tan4 (x7 ) 2x cos 2x + sin 2x + 28x6 tan3 (x7 ) sec2 (x7 ) √ √ 2+5 (7 − x) 3x2 + 5 (7 − x) 3x sec2 2 + 2+ x3 + sin x x3 + sin x

42.

= 17

1 + x2 1 − x2

4x 68x(1 + x2 )16 = (1 − x2 )2 (1 − x2 )18

= 12[1 + sin3 (x5 )]11 3 sin2 (x5 )

41.

dy (4x2 − 1)8 (3)(2x + 3)2 (2) − (2x + 3)3 (8)(4x2 − 1)7 (8x) = dx (4x2 − 1)16 =

40.

1 + x2 1 − x2

39.

1 + x2 1 − x2

Chapter 3

us uf

√

45.

dy dy = 3x2 cos(1 + x3 ); if x = −3 then y = − sin 26, = −27 cos 26, so the equation of the tangent dx dx line is y + sin 26 = −27(cos 26)(x + 3), y = −27(cos 26)x − 81 cos 26 − sin 26

44.

dy dy = cos 3x − 3x sin 3x; if x = π then = −1 and y = −π, so the equation of the tangent line dx dx is y + π = −(x − π), y = x

dy dy = −3 sec3 (π/2 − x) tan(π/2 − x); if x = −π/2 then = 0, y = −1 so the equation of the dx dx tangent line is y + 1 = 0, y = −1

43.

√ 3x2 + 5 (7 − x) 3x2 + 5 (3x2 + cos x) (7 − x)x × − 3 − + 3√ x + sin x (x3 + sin x)2 3x2 + 5 (x3 + sin x)

uh a

2 1 27 dy dy 1 95 135 1 + 2 ; if x = 2 then y = 46. = 3 x− , = 3 = so the equation of the dx x x 8 dx 44 16 108 135 x− tangent line is y − 27/8 = (135/16)(x − 2), y = 8 16

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Exercise Set 3.5

√ √ √ d dy dy = sec2 (4x2 ) (4x2 ) = 8x sec2 (4x2 ), √ = 8 π sec2 (4π) = 8 π. When x = π, dx dx dx x= π √ √ √ y = tan(4π) = 0, so the equation of the tangent line is y = 8 π(x − π) = 8 πx − 8π.

48.

dy d dy = 12 cot3 x cot x = −12 cot3 x csc2 x, = −24. When x = π/4, y = 3, so the dx dx dx x=π/4 equation of the tangent line is y − 3 = −24(x − π/4), or y = −24x + 3 + 6π.

dy x2 dy = 2x 5 − x2 + √ (−2x), = 4 − 1/2 = 7/2. When x = 1, y = 2, so the equation dx dx x=1 2 5 − x2 7 3 of the tangent line is y − 2 = (7/2)(x − 1), or y = x − . 2 2

49.

us uf

47.

dy x dy 1 = 1. When x = 0, y = 0, so the equation of the − (1 − x2 )3/2 (−2x), = √ dx x=0 dx 2 1 − x2 tangent line is y = x.

51.

d d dy = x(− sin(5x)) (5x) + cos(5x) − 2 sin x (sin x) dx dx dx

50.

= −5x sin(5x) + cos(5x) − 2 sin x cos x = −5x sin(5x) + cos(5x) − sin(2x), d d2 y d d = −5x cos(5x) (5x) − 5 sin(5x) − sin(5x) (5x) − cos(2x) (2x) dx2 dx dx dx

52.

d dy = cos(3x2 ) (3x2 ) = 6x cos(3x2 ), dx dx

= −25x cos(5x) − 10 sin(5x) − 2 cos(2x)

d d2 y = 6x(− sin(3x2 )) (3x2 ) + 6 cos(3x2 ) = −36x2 sin(3x2 ) + 6 cos(3x2 ) 2 dx dx (1 − x) + (1 + x) 2 d2 y dy −2 = = = 2(1 − x) and = −2(2)(−1)(1 − x)−3 = 4(1 − x)−3 dx (1 − x)2 (1 − x)2 dx2

54.

dy = x sec2 dx

53.

so y −

1 1 1 1 d 1 1 2 + tan = − sec + tan , x dx x x x x x d2 y 1 2 1 1 1 d 1 1 2 1 d 1 2 2 2 + 2 sec + sec = 3 sec tan = − sec sec dx2 x x dx x x x x dx x x x x 135 27 27 135 (x − 2), y = x− = 2 8 16 16

ad − bc (cu + d)2

d [a cos2 πω + b sin2 πω] = −2πa cos πω sin πω + 2πb sin πω cos πω dω = π(b − a)(2 sin πω cos πω) = π(b − a) sin 2πω

uh a

57.

au + b cu + d

56. 6

55. y = cot3 (π − θ) = − cot3 θ so dy/dx = 3 cot2 θ csc2 θ

58. 2 csc2 (π/3 − y) cot(π/3 − y)

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Chapter 3

59. (a)

−x 4 − 2x2 + 4 − x2 = √ 2 4−x 4 − x2

2 -2

-2

√

-6

(d) f (1) =

us uf

-2

√ 2 2 3 and f (1) = √ so the tangent line has the equation y − 3 = √ (x − 1). 3 3

2 0

60. (a)

0.5

1.2

-1.2

(d) f (1) = sin 1 cos 1 and f (1) = 2 cos2 1 − sin2 1, so the tangent line has the equation y − sin 1 cos 1 = (2 cos2 1 − sin2 1)(x − 1).

0.8

uh a

61. (a) dy/dt = −Aω sin ωt, d2 y/dt2 = −Aω 2 cos ωt = −ω 2 y (b) one complete oscillation occurs when ωt increases over an interval of length 2π, or if t increases over an interval of length 2π/ω (c) f = 1/T (d) amplitude = 0.6 cm, T = 2π/15 s/oscillation, f = 15/(2π) oscillations/s

62. dy/dt = 3A cos 3t, d2 y/dt2 = −9A sin 3t, so −9A sin 3t + 2A sin 3t = 4 sin 3t, −7A sin 3t = 4 sin 3t, −7A = 4, A = −4/7

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Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Exercise Set 3.5

63. (a) p ≈ 10 lb/in2 , dp/dh ≈ −2 lb/in2 /mi

dp dh dp = ≈ (−2)(0.3) = −0.6 lb/in2 /s dt dh dt

us uf

(b)

45 dF 45(− sin θ + 0.3 cos θ) ; , =− cos θ + 0.3 sin θ dθ (cos θ + 0.3 sin θ)2

64. (a) F =

if θ = 30◦ , then dF/dθ ≈ 10.5 lb/rad ≈ 0.18 lb/deg dF dθ dF ≈ (0.18)(−0.5) = −0.09 lb/s = dt dθ dt

66.

d du d d d cos x, u > 0 (|u|) = (|u|) = (|u|) cos x = (| sin x|) = − cos x, u < 0 dx dx du dx du cos x, sin x > 0 cos x, 0<x<π = = − cos x, sin x < 0 − cos x, −π < x < 0

65. With u = sin x,

d d (cos x) = [sin(π/2 − x)] = − cos(π/2 − x) = − sin x dx dx

(b)

67. (a) For x = 0, |f (x)| ≤ |x|, and lim |x| = 0, so by the Squeezing Theorem, lim f (x) = 0. x→0

x→0

f (x) − f (0) (b) If f (0) were to exist, then the limit = sin(1/x) would have to exist, but it x−0 doesn’t. 1 1 1 1 1 1 − 2 + sin = − cos + sin (c) for x = 0, f (x) = x cos x x x x x x f (x) − f (0) = lim sin(1/x), which does not exist, thus f (0) does not exist. x→0 x→0 x−0

(d) lim

68. (a) −x2 ≤ x2 sin(1/x) ≤ x2 , so by the Squeezing Theorem lim f (x) = 0.

x→0

(b) f (0) = lim

x→0

f (x) − f (0) = lim x sin(1/x) = 0 by Exercise 67, Part a. x→0 x−0

(d) If f (x) were continuous at x = 0 then so would cos(1/x) = f (x) − 2x sin(1/x) be, since 2x sin(1/x) is continuous there. But cos(1/x) oscillates at x = 0.

69. (a) g (x) = 3[f (x)]2 f (x), g (2) = 3[f (2)]2 f (2) = 3(1)2 (7) = 21 (b) h (x) = f (x3 )(3x2 ), h (2) = f (8)(12) = (−3)(12) = −36 70. F (x) = f (g(x))g (x) =

√ 3(x2 − 1) + 4(2x) = 2x 3x2 + 1

√ √ 3 3x − 1 3 1 √ 71. F (x) = f (g(x))g (x) = f ( 3x − 1) √ = = (3x − 1) + 1 2 3x − 1 2x 2 3x − 1

uh a

72.

73.

d [f (x2 )] = f (x2 )(2x), thus f (x2 )(2x) = x2 so f (x2 ) = x/2 if x = 0 dx d 2 d [f (3x)] = f (3x) (3x) = 3f (3x) = 6x, so f (3x) = 2x. Let u = 3x to get f (u) = u; dx dx 3 2 d [f (x)] = f (x) = x. dx 3

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Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Chapter 3

f is odd.

d d [f (−x)] = [f (x)], f (−x)(−1) = f (x), f (−x) = −f (x) so dx dx

(b) If f (−x) = −f (x), then f is even.

d d [f (−x)] = − [f (x)], f (−x)(−1) = −f (x), f (−x) = f (x) so dx dx

74. (a) If f (−x) = f (x), then

us uf

f(x)

f '(x) x

75. For an even function, the graph is symmetric about the y-axis; the slope of the tangent line at (a, f (a)) is the negative of the slope of the tangent line at (−a, f (−a)). For an odd function, the graph is symmetric about the origin; the slope of the tangent line at (a, f (a)) is the same as the slope of the tangent line at (−a, f (−a)).

76.

dy du dv dw dy = dx du dv dw dx

77.

d d [f (g(h(x)))] = [f (g(u))], dx dx

EXERCISE SET 3.6 1. y = (2x − 5)1/3 ; dy/dx =

du du d [f (g(u))] = f (g(u))g (u) = f (g(h(x)))g (h(x))h (x) du dx dx

u = h(x)

f '(x)

2 (2x − 5)−2/3 3

−2/3

−2/3 2 1 2 + tan(x2 ) sec2 (x2 )(2x) = x sec2 (x2 ) 2 + tan(x2 ) 3 3

3. dy/dx =

1/2 1/2 9 x−1 d x−1 3 x−1 = 2 x+2 dx x + 2 2(x + 2)2 x + 2

2. dy/dx =

−1/2 −1/2 −1/2 2 1 x2 + 1 x +1 d x2 + 1 −12x 6x 1 x2 + 1 = =− 2 4. dy/dx = 2 x2 − 5 dx x2 − 5 2 x2 − 5 (x2 − 5)2 (x − 5)2 x2 − 5

uh a

2 1 (5x2 + 1)−5/3 (10x) + 3x2 (5x2 + 1)−2/3 = x2 (5x2 + 1)−5/3 (25x2 + 9) 5. dy/dx = x3 − 3 3

4 x2 (3 − 2x)1/3 (−2) − (3 − 2x)4/3 (2x) 2(3 − 2x)1/3 (2x − 9) 3 = 6. dy/dx = x4 3x3

7. dy/dx =

5 15[sin(3/x)]3/2 cos(3/x) [sin(3/x)]3/2 [cos(3/x)](−3/x2 ) = − 2 2x2

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Exercise Set 3.6

8. dy/dx = −

−3/2

−3/2 1 3 cos(x3 ) − sin(x3 ) (3x2 ) = x2 sin(x3 ) cos(x3 ) 2 2

dy dy 2 − 3x2 − y + y − 2 = 0, = dx dx x dy 1 + 2x − x3 1 1 (b) y = = + 2 − x2 so = − 2 − 2x x x dx x dy 2 − 3x2 − y 2 − 3x2 − (1/x + 2 − x2 ) 1 (c) from Part (a), = = = −2x − 2 dx x x x

us uf

dy x2 − 2y dy dy dy = 6(x + y), −(3y 2 + 6x) = 6y − 3x2 so = 2 dx dx dx dx y + 2x

12. 3x2 − 3y 2

x dy dy =− = 0 so dx y dx

dy 1 −1/2 dy √ y − ex = 0 or = 2ex y 2 dx dx dy x x 2 (b) y = (2 + e ) = 2 + 4e + e2x so = 4ex + 2e2x dx dy √ (c) from Part (a), = 2ex y = 2ex (2 + ex ) = 4ex + 2e2x dx

10. (a)

9. (a) 3x2 + x

11. 2x + 2y

dy dy + 2xy + 3x(3y 2 ) + 3y 3 − 1 = 0 dx dx dy dy 1 − 2xy − 3y 3 (x2 + 9xy 2 ) = 1 − 2xy − 3y 3 so = dx dx x2 + 9xy 2

13. x2

dy dy + 3x2 y 2 − 5x2 − 10xy + 1 = 0 dx dx dy dy 10xy − 3x2 y 2 − 1 = 10xy − 3x2 y 2 − 1 so = (2x3 y − 5x2 ) dx dx 2x3 y − 5x2

14. x3 (2y)

15. −

1 y2 dy 1 dy − = − = 0 so y 2 dx x2 dx x2

(x − y)(1 + dy/dx) − (x + y)(1 − dy/dx) , (x − y)2 dy dy x(x − y)2 + y 2x(x − y)2 = −2y + 2x so = dx dx x 1 − 2xy 2 cos(x2 y 2 ) dy dy 17. cos(x2 y 2 ) x2 (2y) + 2xy 2 = 1, = dx dx 2x2 y cos(x2 y 2 )

16. 2x =

(1 + csc y)(− csc2 y)(dy/dx) − (cot y)(− csc y cot y)(dy/dx) , (1 + csc y)2 dy 2x(1 + csc y)2 = − csc y(csc y + csc2 y − cot2 y) , dx 2x(1 + csc y) dy =− but csc2 y − cot2 y = 1, so dx csc y dy dy =1 19. 3 tan2 (xy 2 + y) sec2 (xy 2 + y) 2xy + y2 + dx dx

uh a

18. 2x =

dy 1 − 3y 2 tan2 (xy 2 + y) sec2 (xy 2 + y) = dx 3(2xy + 1) tan2 (xy 2 + y) sec2 (xy 2 + y)

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22.

dy (1 + sec y)[3xy 2 (dy/dx) + y 3 ] − xy 3 (sec y tan y)(dy/dx) = 4y 3 , 2 (1 + sec y) dx dy 2 to get multiply through by (1 + sec y) and solve for dx dy y(1 + sec y) = dx 4y(1 + sec y)2 − 3x(1 + sec y) + xy sec y tan y

us uf

21.

Chapter 3

3x d2 y dy (4y)(3) − (3x)(4dy/dx) 12y − 12x(3x/(4y)) 12y 2 − 9x2 −3(3x2 − 4y 2 ) = , = = = = , 16y 2 16y 2 16y 3 16y 3 dx 4y dx2 d2 y −3(7) 21 but 3x2 − 4y 2 = 7 so = =− 2 3 dx 16y 16y 3

20.

Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

y 2 (2x) − x2 (2ydy/dx) 2xy 2 − 2x2 y(−x2 /y 2 ) 2x(y 3 + x3 ) dy x2 d2 y =− =− =− , = − 2, 2 4 4 y5 dx y dx y y d2 y 2x but x3 + y 3 = 1 so =− 5 dx2 y 2y x(dy/dx) − y(1) x(−y/x) − y dy y d2 y = 2 =− =− =− , dx x dx2 x2 x2 x

24.

y dy = , dx y−x

(y − x)

y y −y −1 y−x y−x (y − x)2

d2 y (y − x)(dy/dx) − y(dy/dx − 1) = = dx2 (y − x)2

23.

y 2 − 2xy d2 y 3 2 but y − 2xy = −3, so =− 3 2 (y − x) dx (y − x)3

sin y d2 y dy dy = = (1 + cos y)−1 , = −(1 + cos y)−2 (− sin y) dx2 dx (1 + cos y)3 dx

26.

dy cos y = , dx 1 + x sin y

25.

(1 + x sin y)(− sin y)(dy/dx) − (cos y)[(x cos y)(dy/dx) + sin y] d2 y = dx2 (1 + x sin y)2 2 sin y cos y + (x cos y)(2 sin2 y + cos2 y) , (1 + x sin y)3

=−

but x cos y = y, 2 sin y cos y = sin 2y, and sin2 y + cos2 y = 1 so

d2 y sin 2y + y(sin2 y + 1) =− 2 (1 + x sin y)3 dx

√ √ x dy dy = − ; at (1/ 2, 1/ 2), = −1; at dx y dx √ √ √ dy dy −x (1/ 2, −1/ 2), = −1 and at = +1. Directly, at the upper point y = 1 − x2 , =√ dx dx 1 − x2 √ x dy the lower point y = − 1 − x2 , =√ = +1. dx 1 − x2

uh a

27. By implicit diﬀerentiation, 2x + 2y(dy/dx) = 0,

√ √ 28. If y 2 − x + 1 = 0, then y = x − 1 goes through the point (10, 3) so dy/dx = 1/(2 x − √ 1). By implicit diﬀerentiation dy/dx = 1/(2y). In both cases, dy/dx|(10,3) = 1/6. Similarly y = − x − 1 √ goes through (10, −3) so dy/dx = −1/(2 x − 1) = −1/6 which yields dy/dx = 1/(2y) = −1/6.

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Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Exercise Set 3.6

dy dy y+1 dy dy + x2 + 2xy + 2x â&#x2C6;&#x2019; 6y = 0, so = â&#x2C6;&#x2019;2x 2 = 0 at x = 0 dx dx dx dx 3y + x2 â&#x2C6;&#x2019; 6y

us uf

30. 3y 2

dy 1 dy x3 = 0, so = â&#x2C6;&#x2019; 3 = â&#x2C6;&#x2019; 3/4 â&#x2030;&#x2C6; â&#x2C6;&#x2019;0.1312. dx dx y 15

â&#x2C6;&#x2019;1/3

34. (a)

â&#x2C6;&#x2019;1/3 dy

= 0,

â&#x2C6;&#x161; â&#x2C6;&#x161; dy y 1/3 = â&#x2C6;&#x2019; 1/3 = 3 at (â&#x2C6;&#x2019;1, 3 3) dx x

y 2

32.

dy dy = 25 2x â&#x2C6;&#x2019; 2y , 31. 4(x2 + y 2 ) 2x + 2y dx dx dy x[25 â&#x2C6;&#x2019; 4(x2 + y 2 )] dy = = â&#x2C6;&#x2019;9/13 ; at (3, 1) dx y[25 + 4(x2 + y 2 )] dx 2 3

29. 4x3 + 4y 3

x 1

â&#x20AC;&#x201C;2

dy dy = (x â&#x2C6;&#x2019; a)(x â&#x2C6;&#x2019; b) + x(x â&#x2C6;&#x2019; b) + x(x â&#x2C6;&#x2019; a) = 3x2 â&#x2C6;&#x2019; 2(a + b)x + ab. If = 0 then dx dx 2 3x â&#x2C6;&#x2019; 2(a + b)x + ab = 0. By the Quadratic Formula 2(a + b) Âą 4(a + b)2 â&#x2C6;&#x2019; 4 Âˇ 3ab 1 a + b Âą (a2 + b2 â&#x2C6;&#x2019; ab)1/2 . x= = 3 6 (c) y = Âą x(x â&#x2C6;&#x2019; a)(x â&#x2C6;&#x2019; b). The square root is only deďŹ ned for nonnegative arguments, so it is necessary that all three of the factors x, x â&#x2C6;&#x2019; a, x â&#x2C6;&#x2019; b be nonnegative, or that two of them be nonpositive. If, for example, 0 < a < b then the function is deďŹ ned on the disjoint intervals 0 < x < a and b < x < +â&#x2C6;&#x17E;, so there are two parts.

(b) 2y

35. (a)

(b) Âą1.1547

-2

â&#x20AC;&#x201C;2

dy dy dy y â&#x2C6;&#x2019; 2x dy â&#x2C6;&#x2019; y + 2y = 0. Solve for = . If =0 dx dx dx 2y â&#x2C6;&#x2019; x dx 2 then y â&#x2C6;&#x2019; 2x = 0 or y = 2x. Thus 4 = x2 â&#x2C6;&#x2019; xy + y 2 = x2 â&#x2C6;&#x2019; 2x2 + 4x2 = 3x2 , x = Âą â&#x2C6;&#x161; . 3

uh a

36.

â&#x2C6;&#x161; 1 â&#x2C6;&#x2019;1/2 du 1 â&#x2C6;&#x2019;1/2 u du u + v = â&#x2C6;&#x2019;â&#x2C6;&#x161; = 0 so 2 dv 2 dv v

37. 4a3

da da da da 2t3 + 3a2 , solve for â&#x2C6;&#x2019; 4t3 = 6 a2 + 2at to get = 3 dt dt dt dt 2a â&#x2C6;&#x2019; 6at

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100

Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Chapter 3

38. 1 = (cos x)

dx dx 1 so = = sec x dy dy cos x

39. 2a2 ω

dω dω b2 λ + 2b2 λ = 0 so =− 2 dλ dλ a ω

us uf

40. Let P (x0 , y0 ) be the required point. The slope of the line 4x − 3y + 1 = 0 is 4/3 so the slope of the tangent to y 2 = 2x3 at P must be −3/4. By implicit diﬀerentiation dy/dx = 3x2 /y, so at P , 3x20 /y0 = −3/4, or y0 = −4x20 . But y02 = 2x30 because P is on the curve y 2 = 2x3 . Elimination of y0 gives 16x40 = 2x30 , x30 (8x0 − 1) = 0, so x0 = 0 or 1/8. From y0 = −4x20 it follows that y0 = 0 when x0 = 0, and y0 = −1/16 when x0 = 1/8. It does not follow, however, that (0, 0) is a solution because dy/dx = 3x2 /y (the slope of the curve as determined by implicit diﬀerentiation) is valid only if y = 0. Further analysis shows that the curve is tangent to the x-axis at (0, 0), so the point (1/8, −1/16) is the only solution.

41. The point (1,1) is on the graph, so 1 + a = b. The slope of the tangent line at (1,1) is −4/3; use 2 4 dy 2xy so at (1,1), − = − , 1 + 2a = 3/2, a = 1/4 implicit diﬀerentiation to get =− 2 1 + 2a 3 dx x + 2ay and hence b = 1 + 1/4 = 5/4. 42. Use implicit diﬀerentiation to get dy/dx = (y − 3x2 )/(3y 2 − x), so dy/dx = 0 if y = 3x2 . Substitute √ √ 3 3 6 3 3 3 this into x − xy + y = 0 to obtain 27x − 2x = 0, x = 2/27, x = 2/3 and hence y = 3 4/3.

43. Let P (x0 , y0 ) be a point where a line through the origin is tangent to the curve x2 − 4x + y 2 + 3 = 0. Implicit diﬀerentiation applied to the equation of the curve gives dy/dx = (2 − x)/y. At P the slope of the curve must equal the slope of the line so (2 − x0 )/y0 = y0 /x0 , or y02 = 2x0 − x20 . But x20 − 4x0 + y02 + 3 = 0 because (x0 , y0 ) is on the curve, − x20 ) + 3 = 0, x0 = 3/2 and elimination of y02 in the latter two equations gives x20 − 4x0 + (2x0 √ 2 2 2 which when√substituted into √ y0 = 2x0 − x0 yields y0 = 3/4, so√y0 = ± 3/2. The √slopes of the lines are (± 3/2)/(3/2) = ± 3/3 and their equations are y = ( 3/3)x and y = −( 3/3)x.

44. By implicit diﬀerentiation, dy/dx = k/(2y) so the slope of the tangent to y 2 = kx at (x0 , y0 ) is k (x − x0 ), or 2y0 y − 2y02 = kx − kx0 . k/(2y0 ) if y0 = 0. The tangent line in this case is y − y0 = 2y0 But y02 = kx0 because (x0 , y0 ) is on the curve y 2 = kx, so the equation of the tangent line becomes 2y0 y − 2kx0 = kx − kx0 which gives y0 y = k(x + x0 )/2. If y0 = 0, then x0 = 0; the graph of y 2 = kx has a vertical tangent at (0, 0) so its equation is x = 0, but y0 y = k(x + x0 )/2 gives the same result when x0 = y0 = 0. dy dy dt = . Use implicit diﬀerentiation on 2y 3 t + t3 y = 1 to get dx dt dx

45. By the chain rule,

dx dy dx dy 3 cos 3x − y 2 dx = y2 = 3(cos 3x) , = dt dt dt dt 2xy dt

uh a

47. 2xy

dx dy dy dy 3x2 y 2 dx + 3x2 y 2 + = 0, =− 3 dt dt dt dt 2x y + 1 dt

46. 2x3 y

2y 3 + 3t2 y 1 2y 3 + 3t2 y dt dy dy =− = =− , but . so 2 3 dt 6ty + t dx cos t dx (6ty 2 + t3 ) cos t

48. (a) f (x) =

(b) f (x) =

4 1/3 4 x , f (x) = x−2/3 3 9 7 4/3 28 1/3 28 −2/3 x , f (x) = x , f (x) = x 3 9 27

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Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Exercise Set 3.7

101

us uf

50. y = rxr−1 , y = r(r − 1)xr−2 so 16x2 r(r − 1)xr−2 + 24x rxr−1 + xr = 0, 16r(r − 1)xr + 24rxr + xr = 0, (16r2 + 8r + 1)xr = 0, 16r2 + 8r + 1 = 0, (4r + 1)2 = 0; r = −1/4

49. y = rxr−1 , y = r(r − 1)xr−2 so 3x2 r(r − 1)xr−2 + 4x rxr−1 − 2xr = 0, 3r(r − 1)xr + 4rxr − 2xr = 0, (3r2 + r − 2)xr = 0, 3r2 + r − 2 = 0, (3r − 2)(r + 1) = 0; r = −1, 2/3

51. We shall ﬁnd when the curves intersect and check that the slopes are negative reciprocals. For the intersection solve the simultaneous equations x2 + (y − c)2 = c2 and (x − k)2 + y 2 = k 2 to obtain y−c 1 x−k cy = kx = (x2 + y 2 ). Thus x2 + y 2 = cy + kx, or y 2 − cy = −x2 + kx, and =− . 2 x y x dy x−k dy =− Diﬀerentiating the two families yields (black) =− . But it was , and (gray) dx y y−c dx proven that these quantities are negative reciprocals of each other.

dy dy + y = 0 and (gray) 2x − 2y = 0. The ﬁrst 52. Diﬀerentiating, we get the equations (black) x dx dx x y says the (black) slope is = − and the second says the (gray) slope is , and these are negative x y reciprocals of each other.

EXERCISE SET 3.7

dy = 3(2) = 6 dt

dy dx +4 =0 dt dt (a) 1 + 4

(a)

dy dx =3 dt dt

dy dy 1 = 0 so = − when x = 2. dt dt 4

dy dx + 2y =0 dt dt √ 3 dy 1 dy 1 (a) 2 (1) + 2 = 0, so = −√ . 2 2 dt dt 3

(b) −1 = 3

(b)

dx dx 1 , =− dt dt 3

dx dx + 4(4) = 0 so = −16 when x = 3. dt dt

3. 2x

√ 2 dx 2 dx (b) 2 +2 (−2) = 0, so = 2. 2 dt 2 dt

dy dx dx + 2y =2 dt dt dt

4. 2x

√

dy dy = −2, =0 dt dt √ √ √ √ dx 2 dx dx 2 + 2 dx + (3) = , (2 + 2 − 2) + 3 2 = 0, = −3 (b) 2 dt 2 dt dt dt

uh a

(a) −2 + 2

5. (b) A = x2 Find

dx dA given that dt x=3 dt x=3

dA dx = 2x dt dt dA = 2. From Part (c), = 2(3)(2) = 12 ft2 /min. dt x=3 (c)

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Chapter 3

6. (b) A = πr2 dr dA given that dt r=5 dt r=5

r=5

(d) Find

dr dA = 2πr dt dt dA = 2. From Part (c), = 2π(5)(2) = 20π cm2 /s. dt (c)

us uf

102

dr dV 2 dh + 2rh . =π r 7. (a) V = πr h, so dt dt dt dV dh dr (b) Find given that = 1 and = −1. From Part (a), dt h=6, dt h=6, dt h=6, r=10 r=10 r=10 dV = π[102 (1) + 2(10)(6)(−1)] = −20π in3 /s; the volume is decreasing. dt h=6,

r=10

1 d+ = dt +

dx dy . +y dt dt

8. (a) +2 = x2 + y 2 , so

d+ 1 dy dx 1 (b) Find = and given that =− . dt 2 dt 4 dt x=3,

y=4

From Part (a) and the fact that + = 5 when x = 3 and y = 4, 1 1 1 d+ 1 3 + 4 − = ft/s; the diagonal is increasing. = dt x=3, 5 2 4 10 y=4

y=2

dx dy −y x y cos2 θ dx dy dθ 2 dθ dt dt 9. (a) tan θ = , so sec θ = = −y x , x dt x2 dt x2 dt dt dθ dx dy 1 (b) Find given that = 1 and =− . dt x=2, dt x=2, dt x=2, 4 y=2 y=2 y=2 π 1 π When x = 2 and y = 2, tan θ = 2/2 = 1 so θ = and cos θ = cos = √ . Thus 4 4 2 √ 5 1 dθ (1/ 2)2 − 2(1) = − rad/s; θ is decreasing. 2 − from Part (a), = dt x=2, 22 4 16

dz dx dy 10. Find given that = −2 and = 3. dt x=1, dt x=1, dt x=1, y=2 y=2 y=2 dz dx dz dy = 2x3 y = (4)(3) + (12)(−2) = −12 units/s; z is decreasing + 3x2 y 2 , dt dt dt dt x=1, y=2

11. Let A be the area swept out, and θ the angle through which the minute hand has rotated. dA dA dθ π 1 dθ 4π 2 Find given that = rad/min; A = r2 θ = 8θ, so =8 = in /min. dt dt 30 2 dt dt 15

uh a

dr dA = 3. given that 12. Let r be the radius and A the area enclosed by the ripple. We want dt t=10 dt dA dr We know that A = πr2 , so = 2πr . Because r is increasing at the constant rate of 3 ft/s, it dt dt dA = 2π(30)(3) = 180π ft2 /s. follows that r = 30 ft after 10 seconds so dt t=10

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Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Exercise Set 3.7

103

us uf

dA dA dr dr dr 1 dA given that 13. Find = 6. From A = πr2 we get = 2πr so = . If A = 9 dt dt dt dt 2πr dt dt A=9 √ √ dr 1 √ (6) = 1/ π mi/h. = then πr2 = 9, r = 3/ π so dt A=9 2π(3/ π)

D 4 where D 14. The volume V of a sphere of radius r is given by V = πr3 or, because r = 3 2 3 4 1 D dD dV 1 is the diameter, V = π = 3. From V = πD3 given that = πD3 . We want 2 6 dt r=1 dt 6 3 dD 1 dD dD 2 3 dV 2 dV ft/min. , , so = πD2 = (3) = we get = 2 dt r=1 π(2)2 2π 2 dt dt πD dt dt

16. Let x and y be the distances shown in the diagram. dx dy given that = 5. From We want to ﬁnd dt y=8 dt

dV 4 dr dV dr 15. Find = −15. From V = πr3 we get = 4πr2 given that so dt dt r=9 dt 3 dt dV = 4π(9)2 (−15) = −4860π. Air must be removed at the rate of 4860π cm3 /min. dt r=9

dx dy dy x dx +2y = 0, so =− . dt dt dt y dt

x2 +y 2 = 172 we get 2x

When y = 8, x2 + 82 = 172 , x2 = 289 − 64 = 225, 15 dy 75 = − (5) = − x = 15 so ft/s; the top of 8 8 dt

y=8

dx dy given that = −2. From x2 + y 2 = 132 dt y=5 dt

we get 2x

dx dy dx y dy + 2y = 0 so = − . dt dt dt x dt

17. Find

the ladder is moving down the wall at a rate of 75/8 ft/s.

Use

x2 + y 2 = 169 to ﬁnd that x = 12 when y = 5 so dx 5 5 = − (−2) = ft/s. dt y=5 12 6

uh a

dθ 18. Let θ be the acute angle, and x the distance of the bottom of the plank from the wall. Find dt x=2 x dx 1 so given that = − ft/s. The variables θ and x are related by the equation cos θ = dt x=2 2 10 √ √ dx dθ 1 dx dθ 1 = , =− . When x = 2, the top of the plank is 102 − 22 = 96 ft dt 10 dt dt 10 sin θ dt √ 1 dθ 1 1 = √ ≈ 0.051 rad/s. = −√ − above the ground so sin θ = 96/10 and dt x=2 2 96 2 96

− sin θ

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104

Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Chapter 3

19. Let x denote the distance from ﬁrst base and y the distance from home plate. Then x2 + 602 = y 2 and √ dy dx = 2y . When x = 50 then y = 10 61 so 2x dt dt x dx dy 125 50 = = √ (25) = √ ft/s. dt y dt 10 61 61

us uf

First

Home

dy dx given that = 2000. From dt x=4 dt x=4 y dy dx dy dx = x2 + 52 = y 2 we get 2x . = 2y so dt dt dt dt x √ Use x2 + 25 = y 2 to ﬁnd that y = 41 when x = 4 so √ √ 41 dx (2000) = 500 41 mi/h. = 4 dt x=4

20. Find

dy dx dy x dx = 2x so = . dt dt dt y dt If x = 4000, then y = 5000 so dy 4000 (880) = 704 ft/s. = dt 5000

dy dx given that = 880. From dt x=4000 dt x=4000

Radar station

y 2 = x2 + 30002 we get 2y

Rocket

Camera 3000 ft

x=4000

5 mi

21. Find

Rocket

dx dφ 22. Find given that = 0.2. But x = 3000 tan φ so dt φ=π/4 dt φ=π/4 dx π

dφ dx (0.2) = 1200 ft/s. = 3000(sec2 φ) , = 3000 sec2 dt dt 4 dt φ=π/4

23. (a) If x denotes the altitude, then r − x = 3960, the radius of the Earth. θ = 0 at perigee, so r = 4995/1.12 ≈ 4460; the altitude is x = 4460 − 3960 = 500 miles. θ = π at apogee, so r = 4995/0.88 ≈ 5676; the altitude is x = 5676 − 3960 = 1716 miles. (b) If θ = 120◦ , then r = 4995/0.94 ≈ 5314; the altitude is 5314 − 3960 = 1354 miles. The rate of change of the altitude is given by dx dr dr dθ 4995(0.12 sin θ) dθ = = = . dt dt dθ dt (1 + 0.12 cos θ)2 dt

uh a

Use θ = 120◦ and dθ/dt = 2.7◦ /min = (2.7)(π/180) rad/min to get dr/dt ≈ 27.7 mi/min.

24. (a) Let x be the horizontal distance shown in the ﬁgure. Then x = 4000 cot θ and

dθ sin2 θ dx dθ dx = −4000 csc2 θ , so =− . Use θ = 30◦ and dt dt dt 4000 dt dx/dt = 300 mi/h = 300(5280/3600) ft/s = 440 ft/s to get

dθ/dt = −0.0275 rad/s ≈ −1.6◦ /s; θ is decreasing at the rate of 1.6◦ /s.

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Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Exercise Set 3.7

105

(b) Let y be the distance between the observation point and the aircraft. Then y = 4000 csc θ so dy/dt = −4000(csc θ cot θ)(dθ/dt). Use θ = 30◦ and dθ/dt = −0.0275 rad/s to get dy/dt ≈ 381 ft/s.

us uf

dh dV = 20. The volume of given that dt h=16 dt 1 water in the tank at a depth h is V = πr2 h. Use 3 10 r 5 similar triangles (see ﬁgure) to get = so r = h h 24 12 2 25 5 dh dV 25 1 = πh2 ; πh3 , h h= thus V = π dt 144 dt 3 12 432 dh 144 9 144 dV dh , = (20) = = 25π(16)2 20π dt 25πh2 dt dt h=16 ft/min.

25. Find

1 dV dh = 8. V = πr2 h, but given that 26. Find dt h=6 dt 3 2 h 1 dh 1 dV 1 1 r = h so V = π h= πh3 , = πh2 , 2 3 2 12 dt 4 dt 4 dV dh 4 8 dh = = (8) = , ft/min. π(6)2 9π dt πh2 dt dt h=6

dV dh 1 given that = 5. V = πr2 h, but dt h=10 dt 3 2 1 dh 1 dV 1 h 1 r = h so V = π h= πh3 , = πh2 , 2 3 2 12 dt 4 dt 1 dV = π(10)2 (5) = 125π ft3 /min. dt h=10 4

27. Find

uh a

28. Let r and h be as shown in the ﬁgure. If C is the circumference dV dC given that of the base, then we want to ﬁnd = 10. dt h=8 dt 1 It is given that r = h, thus C = 2πr = πh so 2 dC dh =π (1) dt dt

Use

V =

1 dh 1 dV 1 2 πr h = πh3 to get = πh2 , so 3 12 dt 4 dt

dh 4 dV = dt πh2 dt

Substitution of (2) into (1) gives

(2) dC 4 4 dV dC 5 = = 2 so (10) = ft/min. dt h dt dt h=8 64 8

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Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Chapter 3

29. With s and h as shown in the figure, we want ds dh given that = 500. From the figure, to find dt dt dh 1 ds 1 1 h = s sin 30◦ = s so = = (500) = 250 mi/h. 2 dt 2 dt 2

30°

us uf

106

Ground

given that y=125

dy = −20. From dt

Pulley

dx dx y dy dy . Use so = = 2y x dt dt dt dt √ √ x2 + 100 = y 2 to ﬁnd that x = 15, 525 = 15 69 when dx 125 500 y = 125 so = √ (−20) = − √ . dt 15 69 3 69 x2 + 102 = y 2 we get 2x

dx 30. Find dt

Boat

y=125

500 The boat is approaching the dock at the rate of √ ft/min. 3 69 dx dy given that 31. Find = −12. From x2 +102 = y 2 dt y=125 dt dx dy dy x dx = 2y so = . Use x2 + 100 = y 2 dt dt dt y dt √ to ﬁnd that x = 15, 525 = 15 69 when y = 125 so √ √ dy 15 69 36 69 = (−12) = − . The rope must be pulled dt 125 25 √ 36 69 ft/min. at the rate of 25

we get 2x

Pulley

Boat

Light

18 Man Shadow

32. (a) Let x and y be as shown in the ﬁgure. It is required dy dx , given that = −3. By similar triangles, to ﬁnd dt dt x x+y 1 = , 18x = 6x + 6y, 12x = 6y, x = y, so 6 18 2 1 dy dx 1 3 = = (−3) = − ft/s. 2 dt 2 2 dt

uh a

(b) The tip of the shadow is z = x + y feet from the street light, thus the rate at which it is dx dz dy dx 3 dy = moving is given by + . In part (a) we found that = − when = −3 so dt dt dt 2 dt dt dz = (−3/2) + (−3) = −9/2 ft/s; the tip of the shadow is moving at the rate of 9/2 ft/s dt toward the street light.

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Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Exercise Set 3.7

dθ dx 2π π given that 33. Find = = rad/s. dt 10 5 dt θ=π/4

107

θ=π/4

us uf

dθ dx = 4 sec2 θ , Then x = 4 tan θ (see ﬁgure) so dt dt

π dx = 8π/5 km/s. = 4 sec2 dt 4 5

Ship

Missile

y=4

Aircraft

34. If x, y, and z are as shown in the figure, then we want dz dx dy = −1200. But given that = −600 and dt x=2, dt dt x=2, y=4 y=4 dy dx dy dz dx dz 1 2 2 2 +y . = 2x + 2y , x z = x + y so 2z = dt dt dt dt dt dt z √ √ When x = 2 and y = 4, z 2 = 22 + 42 = 20, z = 20 = 2 5 so √ 1 3000 dz = √ [2(−600) + 4(−1200)] = − √ = −600 5 mi/h; dt x=2, 2 5 5 the distance √ between missile and aircraft is decreasing at the rate of 600 5 mi/h. dx dz given 35. We wish to find = −600 and dt x=2, dt y=4 dy = −1200 (see figure). From the law of cosines, dt x=2, ◦

y=4

120º z

dx dy dy dx dz = 2x +2y +x +y , dt dt dt dt dt dx dz 1 dy (2x + y) . = + (2y + x) dt 2z dt dt

= x2 +y 2 +xy, so 2z

Aircraft

z = x + y − 2xy cos 120 = x + y − 2xy(−1/2) 2

Missile

√ √ When x = 2 and y = 4, z 2 = 22 + 42 + (2)(4) = 28, so z = 28 = 2 7, thus √ dz 1 4200 √ [(2(2) + 4)(−600) + (2(4) + 2)(−1200)] = − √ = −600 7 mi/h; = dt x=2, 2(2 7) 7 y=4

√ the distance between missile and aircraft is decreasing at the rate of 600 7 mi/h.

uh a

36. (a) Let P be the point on the helicopter’s path that lies directly above the car’s path. Let x, dx dz given that y, and z be the distances shown in the first figure. Find = −75 and dt dt x=2, y=0 dy = 100. In order to find an equation relating x, y, and z, first draw the line segment dt that joins the point P to the car, as shown in the second figure. Because triangle OP C is a right triangle, it follows that P C has length x2 + (1/2)2 ; but triangle HP C is also a right

2 dz dx dy x2 + (1/2)2 + y 2 = x2 + y 2 + 1/4 and 2z triangle so z 2 = = 2x + 2y + 0, dt dt dt dx 1 dy dz x . Now, when x = 2 and y = 0, z 2 = (2)2 + (0)2 + 1/4 = 17/4, = +y z dt dt dt

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108

Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Chapter 3

√

√ 1 dz = √ 17/2 so [2(−75) + 0(100)] = −300/ 17 mi/h dt x=2, ( 17/2) y=0

1 2

x Car

West

East

z Helicopter

(b) decreasing, because

P y

dz < 0. dt

dy dx given that = 6. For convenience, ﬁrst rewrite the equation as dt x=1, dt x=1, y=2

y=2

37. (a) We want

us uf

1 2

North

16 dy dy y3 dy dx dx 8 8 2 + y then 3xy 2 + y3 = y , = so 16 5 5 dt dt 5 dt dt dt y − 3xy 2 5 dy 23 = (6) = −60/7 units/s. 16 dt x=1, y=2 (2) − 3(1)22 5

(b) falling, because

dy <0 dt

xy 3 =

dx dy 38. Find given that = 2. Square and rearrange to get x3 = y 2 − 17 dt (2,5) dt (2,5) 5 dy dx 5 2y dy dx 2 dx (2) = units/s. = 2y , so 3x = = 2 , 3x dt dt (2,5) 6 3 dt dt dt

39. The coordinates of P are (x, 2x), so the distance between P and the point (3, 0) is √ dD dx D = (x − 3)2 + (2x − 0)2 = 5x2 − 6x + 9. Find given that = −2. dt x=3 dt x=3 5x − 3 dD dD 12 dx =√ , so = √ (−2) = −4 units/s. 2 dt dt dt 36 5x − 6x + 9 x=3

uh a

dD dx 40. (a) Let D be the distance between P and (2, 0). Find given that = 4. dt x=3 dt x=3 √ dD 2x − 3 ; D = (x − 2)2 + y 2 = (x − 2)2 + x = x2 − 3x + 4 so = √ dt 2 x2 − 3x + 4 3 dD 3 = √ = units/s. dt x=3 4 2 4 dθ dx (b) Let θ be the angle of inclination. Find given that = 4. dt x=3 dt x=3 √ x dx dθ dx dθ x+2 y x+2 tan θ = = so sec2 θ =− √ , = − cos2 θ √ . x−2 x−2 dt 2 x(x − 2)2 dt dt 2 x(x − 2)2 dt 1 5 5 dθ 1 = − √ (4) = − √ rad/s. When x = 3, D = 2 so cos θ = and 2 dt x=3 42 3 2 3

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Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Exercise Set 3.7

109

dy dx =3 given y = x/(x2 + 1). Then y(x2 + 1) = x. Diﬀerentiating with respect to x, dt dt dy dy/dt 1 dy 1 + y(2x) = 1. But = = so (x2 + 1) + 2xy = 1, x2 + 1 + 6xy = 3, (x2 + 1) dx dx dx/dt 3 3 x2 + 1 + 6x2 /(x2 + 1) = 3, (x2 + 1)2 + √ 6x2 − 3x2 − 3 = 0, x4 + 5x2 − 3 = 0. By the binomial theorem x2 = (−5 ± 25 + 12)/2. The minus sign is spurious since x2 cannot be applied to x2 we obtain √ 2 negative, so x = ( 33 − 5)/2, x ≈ ±0.6101486081, y = ±0.4446235604.

us uf

41. Solve

dy dx dx dx 16 dy + 18y = = 0, then (32x + 18y) = 0, 32x + 18y = 0, y = − x so = 0; if dt dt 9 dt dt dt 9 9 16 9 16 256 2 400 2 81 2 2 x = 144, x = 144, x = , x = ± . If x = , then y = − = − . 16x + 9 81 9 25 5 5 9 5 5 9 16 9 16 16 9 and − , . . The points are ,− Similarly, if x = − , then y = 5 5 5 5 5 5

42. 32x

1 1 1 ds 1 dS dS ds 1 = we get − 2 − 2 = 0, so 43. Find given that = −2. From + s S 6 s dt S dt dt s=10 dt s=10 dS 1 1 dS S 2 ds 1 225 + = which gives S = 15. So (−2) = 4.5 =− 2 . If s = 10, then =− 10 S 6 dt s=10 100 dt s dt cm/s. The image is moving away from the lens.

44. Suppose that the reservoir has height H and that the radius at the top is R. At any instant of time let h and r be the corresponding dimensions of the cone of water (see ﬁgure). We want to show dV dh is constant and independent of H and R, given that = −kA where V is the volume that dt dt of water, A is the area of a circle of radius r, and k is a positive constant. The volume of a cone of 2 1 R R 1 R r radius r and height h is V = πr2 h. By similar triangles = , r = h thus V = π h3 h H H 3 H 3 so 2 R dh dV (1) =π h2 H dt dt dV = −kA or, because dt 2 2 R R 2 2 dV = −kπ h , h2 , A = πr = π H dt H

R r

But it is given that

which when substituted into equation (1) gives 2 2 R dh dh R −kπ h2 = π h2 , = −k. H H dt dt

uh a

dr 45. Let r be the radius, V the volume, and A the surface area of a sphere. Show that is a constant dt 4 3 dV = −kA, where k is a positive constant. Because V = πr , given that dt 3 dV dr = 4πr2 dt dt

(1)

dV dV = −kA or, because A = 4πr2 , = −4πr2 k which when substituted into dt dt dr dr equation (1) gives −4πr2 k = 4πr2 , = −k. dt dt

But it is given that

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Chapter 3

us uf

46. Let x be the distance between the tips of the minute and hour hands, and α and β the angles shown in the figure. Because the minute hand makes one revolution in 60 minutes, dα 2π = π/30 rad/min; the hour hand makes one revolution in 12 hours (720 minutes), thus = 60 dt 2π dx dβ dα dβ = = π/360 rad/min. We want to find = π/30 and = π/360. given that dt 720 dt α=2π, dt dt Using the law of cosines on the triangle shown in the figure, x2 = 32 + 42 − 2(3)(4) cos(α − β) = 25 − 24 cos(α − β), so dα dβ dx = 0 + 24 sin(α − β) − , 2x dt dt dt dx 12 dα dβ sin(α − β). When α = 2π and β = 3π/2, = − x dt dt dt

β=3π/2

x2 = 25 − 24 cos(2π − 3π/2) = 25, x = 5; so 11π dx 12 (π/30 − π/360) sin(2π − 3π/2) = in/min. = dt α=2π, 5 150

47. Extend sides of cup to complete the cone and let V0 be the volume of the portion added, then (see ﬁgure) r 1 4 1 1 V = πr2 h − V0 where = = so r = h and 3 h 12 3 3 2 dh 1 dV h 1 1 h − V0 = πh3 − V0 , = πh2 , V = π 3 3 27 dt 9 dt 9 dV dh 9 20 dh , cm/s. = = (20) = πh2 dt dt π(9)2 9π dt

4 r

6 2 6

h=9

EXERCISE SET 3.8

β=3π/2

␤ ␣

1. (a) f (x) ≈ f (1) + f (1)(x − 1) = 1 + 3(x − 1) (b) f (1 + ∆x) ≈ f (1) + f (1)∆x = 1 + 3∆x (c) From Part (a), (1.02)3 ≈ 1 + 3(0.02) = 1.06. From Part (b), (1.02)3 ≈ 1 + 3(0.02) = 1.06.

2. (a) f (x) ≈ f (2) + f (2)(x − 2) = 1/2 + (−1/22 )(x − 2) = (1/2) − (1/4)(x − 2) (b) f (2 + ∆x) ≈ f (2) + f (2)∆x = 1/2 − (1/4)∆x (c) From Part (a), 1/2.05 ≈ 0.5 − 0.25(0.05) = 0.4875, and from Part (b), 1/2.05 ≈ 0.5 − 0.25(0.05) = 0.4875.

uh a

√ 3. (a) f (x) ≈ f (x0 ) + f (x √0 )(x − x0 ) = 1 + (1/(2 1)(x − 0) = 1 + (1/2)x, so with x0 = 0 and x = −0.1, we have 0.9 = f (−0.1) ≈ 1 + (1/2)(−0.1) = 1 − 0.05 = 0.95. With x = 0.1 we √ have 1.1 = f (0.1) ≈ 1 + (1/2)(0.1) = 1.05.

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Exercise Set 3.8

(b)

111

us uf

∆y

∆y dy

x 0.1

–0.1

y ∆y dy

dy 3.9

∆y

(b)

4. (a) f (x) ≈ f (x0 ) + f (x0 )(x − x0 ) = 1/2 − 1/(2 · 43/2 ) (x − 4) = 1/2 − (x − 4)/16, so with x0 = 4 √ and x =√3.9 we have 1/ 3.9 = f (3.9) ≈ 0.5 − (−0.1)/16 = 0.50625. If x0 = 4 and x = 4.1 then 1/ 4.1 = f (4.1) ≈ 0.5 − (0.1)/16 = 0.49375

4.1

5. f (x) = (1+x)15 and x0 = 0. Thus (1+x)15 ≈ f (x0 )+f (x0 )(x−x0 ) = 1+15(1)14 (x−0) = 1+15x.

7. tan x ≈ tan(0) + sec2 (0)(x − 0) = x

1 1 1 and x0 = 0, so √ (x − 0) = 1 + x/2 ≈ f (x0 ) + f (x0 )(x − x0 ) = 1 + 2(1 − 0)3/2 1−x 1−x 8.

6. f (x) = √

1 −1 (x − 0) = 1 − x ≈1+ 1+x (1 + 0)2

10.

x≈

√

√ 1 1 + √ (x − 1), and x = 1 + ∆x, so 1 + ∆x ≈ 1 + ∆x/2 2 1

1 1 1 1 1 1 (x − 1), and 2 + x = 3 + ∆x, so ≈ − ≈ − ∆x 2 2+x 2 + 1 (2 + 1) 3 + ∆x 3 9

11.

√

9. x4 ≈ (1)4 + 4(1)3 (x − 1). Set ∆x = x − 1; then x = ∆x + 1 and (1 + ∆x)4 = 1 + 4∆x.

12. (4 + x)3 ≈ (4 + 1)3 + 3(4 + 1)2 (x − 1) so, with 4 + x = 5 + ∆x we get (5 + ∆x)3 ≈ 125 + 75∆x -2

-0.1

| f (x) – ( 3 +

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√ 13. f (x) = x + 3 and x0 = 0, so √ √ √ 1 1 x + 3 ≈ 3 + √ (x − 0) = 3 + √ x, and 2 3 2 3 √ 1 f (x) − < 0.1 if |x| < 1.692. √ 3 + x 2 3

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1 2 3

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Chapter 3

1 so 9−x 1 1 1 1 1 √ (x − 0) = + x, ≈√ + 3 54 9−x 9 2(9 − 0)3/2 1 1 + x < 0.1 if |x| < 5.5114 and f (x) − 3 54

14. f (x) = √

us uf

0.06

-6

1 3

15. tan x ≈ tan 0 + (sec2 0)(x − 0) = x, and | tan x − x| < 0.1 if |x| < 0.6316

0.06

-0.8

−5(2) 1 1 + (x − 0) = 1 − 10x, ≈ (1 + 2 · 0)5 (1 + 2 · 0)6 (1 + 2x)5

0.8 0 | f (x) – x |

0.12

16.

1 x 54

| f (x) – (

and |f (x) − (1 − 10x)| < 0.1 if |x| < 0.0372

-0.04

0.04 0 | f (x) – (1 – 10x)|

17. (a) The local linear approximation sin x ≈ x gives sin 1◦ = sin(π/180) ≈ π/180 = 0.0174533 and a calculator gives sin 1◦ = 0.0174524. The relative error | sin(π/180)−(π/180)|/(sin π/180) = 0.000051 is very small, so for such a small value of x the approximation is very good. √ (b) Use x0 = 45◦ (this assumes you know, or can approximate, 2/2). 44π 45π π π 44π (c) 44◦ = radians, and 45◦ = = radians. With x = and x0 = we obtain 180 180 4 180 4 √ √

44π π 2 2 −π 44π π π = = 0.694765. With a ≈ sin + cos − + sin 44◦ = sin 180 4 4 180 4 2 2 180 calculator, sin 44◦ = 0.694658.

uh a

18. (a) tan x ≈ tan 0 + sec2 0(x − 0) = x, so tan 2◦ = tan(2π/180) ≈ 2π/180 = 0.034907, and with a calculator tan 2◦ = 0.034921 √ (b) use x0 = π/3 because we know tan 60◦ = tan(π/3) = 3 60π 61π π and x = we have (c) with x0 = = 3 180 180 √ π 61π π 61π π π = 3+4 tan 61◦ = tan ≈ tan + sec2 − = 1.8019, 180 3 3 180 3 180 and with a calculator tan 61◦ = 1.8040

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Exercise Set 3.8

113

22. f (x) =

23. f (x) =

√ √ 1 1 1 = 8.0625 x, f (x) = √ , x0 = 64, ∆x = 1; 65 ≈ 64 + (1) = 8 + 16 16 2 x

√

√ √ 1 1 x, f (x) = √ , x0 = 25, ∆x = −1; 24 ≈ 25 + (−1) = 5 − 0.1 = 4.9 10 2 x

√

√ √ 1 1 x, f (x) = √ , x0 = 81, ∆x = −0.1; 80.9 ≈ 81 + (−0.1) ≈ 8.9944 18 2 x

√

√ √ 1 1 x, f (x) = √ , x0 = 36, ∆x = 0.03; 36.03 ≈ 36 + (0.03) = 6 + 0.0025 = 6.0025 12 2 x

24. f (x) =

√

21. f (x) =

us uf

20. f (x) = x3 , f (x) = 3x2 , x0 = 2, ∆x = −0.03; (1.97)3 ≈ 23 + (12)(−0.03) = 8 − 0.36 = 7.64

19. f (x) = x4 , f (x) = 4x3 , x0 = 3, ∆x = 0.02; (3.02)4 ≈ 34 + (108)(0.02) = 81 + 2.16 = 83.16

25. f (x) = sin x, f (x) = cos x, x0 = 0, ∆x = 0.1; sin 0.1 ≈ sin 0 + (cos 0)(0.1) = 0.1 26. f (x) = tan x, f (x) = sec2 x, x0 = 0, ∆x = 0.2; tan 0.2 ≈ tan 0 + (sec2 0)(0.2) = 0.2

27. f (x) = cos x, f (x) = − sin x, x0 = π/6, ∆x = π/180;

√3 π 1 π = cos 31◦ ≈ cos 30◦ + − − ≈ 0.8573 2 180 2 360

28. (a) Let f (x) = (1 + x)k and x0 = 0. Then (1 + x)k ≈ 1k + k(1)k−1 (x − 0) = 1 + kx. Set k = 37 and x = 0.001 to obtain (1.001)37 ≈ 1.037. (b) With a calculator (1.001)37 = 1.03767. (c) The approximation is (1.1)37 ≈ 1 + 37(0.1) = 4.7, and the calculator value is 34.004. The error is due to the relative largeness of f (1)∆x = 37(0.1) = 3.7. y

(b)

29. (a) dy = f (x)dx = 2xdx = 4(1) = 4 and ∆y = (x + ∆x)2 − x2 = (2 + 1)2 − 22 = 5

m m

uh a

30. (a) dy = 3x2 dx = 3(1)2 (1) = 3 and ∆y = (x + ∆x)3 − x3 = (1 + 1)3 − 13 = 7

∆y

x 2

(b)

8 ∆y dy 1

(1,1)

dx 1

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Chapter 3

(b) 2

∆y = 1

31. (a) dy = (−1/x2 )dx = (−1)(−0.5) = 0.5 and ∆y = 1/(x + ∆x) − 1/x = 1/(1 − 0.5) − 1/1 = 2 − 1 = 1

dy = 0.5 1

us uf

114

√ 32. (a) dy = (1/2 x)dx = (1/(2 · 3))(−1) = −1/6 ≈ −0.167 and √ √ √ √ ∆y = x + ∆x − x = 9 + (−1) − 9 = 8 − 3 ≈ −0.172 y

(b) 3

dy = –0.167

0.5

∆ y = –0.171

x 8

33. dy = 3x2 dx; ∆y = (x + ∆x)3 − x3 = x3 + 3x2 ∆x + 3x(∆x)2 + (∆x)3 − x3 = 3x2 ∆x + 3x(∆x)2 + (∆x)3 34. dy = 8dx; ∆y = [8(x + ∆x) − 4] − [8x − 4] = 8∆x

35. dy = (2x − 2)dx; ∆y = [(x + ∆x)2 − 2(x + ∆x) + 1] − [x2 − 2x + 1] = x2 + 2x ∆x + (∆x)2 − 2x − 2∆x + 1 − x2 + 2x − 1 = 2x ∆x + (∆x)2 − 2∆x 36. dy = cos x dx; ∆y = sin(x + ∆x) − sin x

37. (a) dy = (12x2 − 14x)dx (b) dy = x d(cos x) + cos x dx = x(− sin x)dx + cos xdx = (−x sin x + cos x)dx

√

x 1−x− √ 2 1−x

39. (a) dy =

38. (a) dy = (−1/x2 )dx

(b) dy = 5 sec2 x dx 2 − 3x dx = √ dx 2 1−x

(b) dy = −17(1 + x)−18 dx

(x3 − 1)(0) − (1)3x2 dx 3x2 (x3 − 1)d(1) − (1)d(x3 − 1) = =− 3 dx 3 2 3 2 (x − 1) (x − 1) (x − 1)2 2x3 − 6x2 + 1 (2 − x)(−3x2 )dx − (1 − x3 )(−1)dx = dx 2 (2 − x) (2 − x)2

uh a

40. (a) dy =

(b) dy =

3 3 41. dy = √ dx, x = 2, dx = 0.03; ∆y ≈ dy = (0.03) = 0.0225 4 2 3x − 2

42. dy = √

x dx, x = 1, dx = −0.03; ∆y ≈ dy = (1/3)(−0.03) = −0.01 x2 + 8

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Exercise Set 3.8

1 − x2 dx, x = 2, dx = −0.04; ∆y ≈ dy = (x2 + 1)2 √

−

3 25

(−0.04) = 0.0048

√ 4x + 8x + 1 dx, x = 3, dx = 0.05; ∆y ≈ dy = (37/5)(0.05) = 0.37 8x + 1

44. dy =

us uf

43. dy =

115

45. (a) A = x2 where x is the length of a side; dA = 2x dx = 2(10)(±0.1) = ±2 ft2 .

dx ±0.1 = ±0.01 so percentage error in x is ≈ ±1%; relative error (b) relative error in x is ≈ = x 10 dA 2x dx dx in A is ≈ = = 2(±0.01) = ±0.02 so percentage error in A is ≈ ±2% =2 A x2 x 46. (a) V = x3 where x is the length of a side; dV = 3x2 dx = 3(25)2 (±1) = ±1875 cm3 .

dx ±1 = = ±0.04 so percentage error in x is ≈ ±4%; relative error x 25 dx 3x2 dx dV = 3(±0.04) = ±0.12 so percentage error in V is ≈ ±12% =3 = in V is ≈ V x3 x

(b) relative error in x is ≈

47. (a) x = 10 sin θ, y = 10 cos θ (see ﬁgure), √ π π

3 π

10″ dx = 10 cos θdθ = 10 cos ± = 10 ± 6 180 2 180 ≈ ±0.151 in, θ π π

1 π

y dy = −10(sin θ)dθ = −10 sin ± = −10 ± 6 180 2 180 ≈ ±0.087 in π π √ π

dx = (cot θ)dθ = cot ± = 3 ± ≈ ±0.030 (b) relative error in x is ≈ x 6 180 180 so percentage error in x is ≈ ±3.0%; π π

1 π

dy = − tan θdθ = − tan ± = −√ ± ≈ ±0.010 relative error in y is ≈ y 6 180 180 3 so percentage error in y is ≈ ±1.0%

25 cm

48. (a) x = 25 cot θ, y = 25 csc θ (see ﬁgure); π π

dx = −25 csc2 θdθ = −25 csc2 ± 3 360

4 π = −25 ± ≈ ±0.291 cm, 3 360 π π π

dy = −25 csc θ cot θdθ = −25 csc cot ± 3 3 360 2 1 π

√ = −25 √ ≈ ±0.145 cm ± 360 3 3

csc2 θ dx 4/3 π

=− ≈ ±0.020 so percentage dθ = − √ ± x cot θ 360 1/ 3 dy 1 π

≈ ±0.005 error in x is ≈ ±2.0%; relative error in y is ≈ = − cot θdθ = − √ ± y 360 3 so percentage error in y is ≈ ±0.5%

uh a

(b) relative error in x is ≈

49.

dR (−2k/r3 )dr dr dr dR = = −2 , but ≈ ±0.05 so ≈ −2(±0.05) = ±0.10; percentage error in R 2 R (k/r ) r r R is ≈ ±10%

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Chapter 3

2x dx dx dA dx = ≈ ±0.01 so = 2 , but 52. A = x2 where x is the length of a side; 2 A x x x dA ≈ 2(±0.01) = ±0.02; percentage error in A is ≈ ±2% A

us uf

1 2 (4) sin 2θ = 4 sin 2θ thus dA = 8 cos 2θdθ so, with θ = 30◦ = π/6 radians and 4 dθ = ±15 = ±1/4◦ = ±π/720 radians, dA = 8 cos(π/3)(±π/720) = ±π/180 ≈ ±0.017 cm2

51. A =

50. h = 12 sin θ thus dh = 12 cos θdθ so, with θ = 60◦ = π/3 radians and dθ = −1◦ = −π/180 radians, dh = 12 cos(π/3)(−π/180) = −π/30 ≈ −0.105 ft

dx dx dV 3x2 dx , but ≈ ±0.02 =3 53. V = x3 where x is the length of a side; = V x3 x x dV so ≈ 3(±0.02) = ±0.06; percentage error in V is ≈ ±6%. V

4πr2 dr dr dV dr dr dV = = 3 , but ≈ ±0.03 so 3 ≈ ±0.03, ≈ ±0.01; maximum permissible V 4πr3 /3 r V r r percentage error in r is ≈ ±1%.

54.

1 (πD/2)dD dD dA dA = =2 , but ≈ ±0.01 so πD2 where D is the diameter of the circle; 2 A πD /4 D A 4 dD dD 2 ≈ ±0.005; maximum permissible percentage error in D is ≈ ±0.5%. ≈ ±0.01, D D

55. A =

56. V = x3 where x is the length of a side; approximate ∆V by dV if x = 1 and dx = ∆x = 0.02, dV = 3x2 dx = 3(1)2 (0.02) = 0.06 in3 .

57. V = volume of cylindrical rod = πr2 h = πr2 (15) = 15πr2 ; approximate ∆V by dV if r = 2.5 and dr = ∆r = 0.001. dV = 30πr dr = 30π(2.5)(0.001) ≈ 0.236 cm3 .

2π √ π dP 2π 1 1 dL 1 58. P = √ L, dP = √ √ dL = √ √ dL, = so the relative error in P ≈ the g g2 L P 2 L 2 g L 1 relative error in L. Thus the percentage error in P is ≈ the percentage error in L. 2

59. (a) α = ∆L/(L∆T ) = 0.006/(40 × 10) = 1.5 × 10−5/◦ C (b) ∆L = 2.3 × 10−5 (180)(25) ≈ 0.1 cm, so the pole is about 180.1 cm long.

60. ∆V = 7.5 × 10−4 (4000)(−20) = −60 gallons; the truck delivers 4000 − 60 = 3940 gallons.

CHAPTER 3 SUPPLEMENTARY EXERCISES

uh a

4. (a)

√ 9 − 4(x + h) − 9 − 4x (9 − 4(x + h) − (9 − 4x) = lim √ h→0 h( 9 − 4(x + h) + h 9 − 4x) −4h −4 −2 = √ = lim =√ √ h→0 h( 9 − 4(x + h) + 2 9 − 4x 9 − 4x 9 − 4x)

dy = lim dx h→0

(b)

x x+h − dy x + 1 = lim (x + h)(x + 1) − x(x + h + 1) x + h + 1 = lim h→0 dx h→0 h h(x + h + 1)(x + 1) h 1 = lim = h→0 h(x + h + 1)(x + 1) (x + 1)2

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Chapter 3 Supplementary Exercises

117

us uf

5. Set f (x) = 0: f (x) = 6(2)(2x + 7)5 (x − 2)5 + 5(2x + 7)6 (x − 2)4 = 0, so 2x + 7 = 0 or x − 2 = 0 or, factoring out (2x + 7)5 (x − 2)4 , 12(x − 2) + 5(2x + 7) = 0. This reduces to x = −7/2, x = 2, or 22x + 11 = 0, so the tangent line is horizontal at x = −7/2, 2, −1/2. 4(x2 + 2x)(x − 3)3 − (2x + 2)(x − 3)4 , and a fraction can equal zero (x2 + 2x)2 only if its numerator equals zero. So either x − 3 = 0 or, after factoring out (x − 3)3 , 4(x2 + 2x) − (2x + 2)(x − 3) = 0, 2x2 + 12x + 6 = 0, whose roots are (by the quadratic formula) √ √ √ −6 ± 36 − 4 · 3 = −3 ± 6. So the tangent line is horizontal at x = 3, −3 ± 6. x= 2

6. Set f (x) = 0: f (x) =

√ 3 (x − 1)2 + 2 3x + 1(x − 1) = 0. If x = 1 then y = 0. If x = 1 then divide 7. Set f (x) = √ 2 3x + 1 √ out x − 1 and multiply through by 2 3x + 1 (at points where f is diﬀerentiable we must have √ 3x + 1 = 0) to obtain 3(x − 1) + 4(3x + 1) = 0, or 15x + 1 = 0. So the tangent line is horizontal at x = 1, −1/15. 2 2 2 d 3x + 1 x (3) − (3x + 1)(2x) 3x + 1 3x + 1 = 3 x2 dx x2 x2 x4 2 2 3x + 2x 3x + 1 = −3 = 0. If f (x) = 0 2 x x4

8. f (x) = 3

x = −2, −1, 1, 3 (−∞, −2), (−1, 1), (3, +∞) (−2, −1), (1, 3) g (x) = f (x) sin x + 2f (x) cos x − f (x) sin x; g (0) = 2f (0) cos 0 = 2(2)(1) = 4

9. (a) (b) (c) (d)

then (3x + 1)2 (3x2 + 2x) = 0. The tangent line is horizontal at x = −1/3, −2/3 (x = 0 is ruled out from the deﬁnition of f ).

10. (a) f (1)g(1) + f (1)g (1) = 3(−2) + 1(−1) = −7

(c)

−2(3) − 1(−1) 5 g(1)f (1) − f (1)g (1) = =− g(1)2 (−2)2 4 1 1 3 f (1) = √ 3 = 2 2 1 2 f (1)

(b)

(d) 0 (because f (1)g (1) is constant)

uh a

11. The equation of such a line has the form y = mx. The points (x0 , y0 ) which lie on both the line and the parabola and for which the slopes of both curves are equal satisfy y0 = mx0 = x30 − 9x20 − 16x0 , so that m = x20 − 9x0 − 16. By diﬀerentiating, the slope is also given by m = 3x20 − 18x0 − 16. Equating, we have x20 − 9x0 − 16 = 3x20 − 18x0 − 16, or 2x20 − 9x0 = 0. The root x0 = 0 corresponds to m = −16, y0 = 0 and the root x0 = 9/2 corresponds to m = −145/4, y0 = −1305/8. So the line y = −16x is tangent to the curve at the point (0, 0), and the line y = −145x/4 is tangent to the curve at the point (9/2, −1305/8).

12. The slope of the line x + 4y = 10 is m1 = −1/4, so we set the negative reciprocal d 4 = m2 = (2x3 − x2 ) = 6x2 − 2x and obtain 6x2 − 2x − 4 = 0 with roots dx √ 1 ± 1 + 24 x= = 1, −2/3. 6

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118

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Chapter 3

(a) (b)

us uf

14. f (x) is continuous and diﬀerentiable at any x = 1, so we consider x = 1.

d (3x − tan x) = 3 − sec2 x = 1, or sec2 x = 2, 13. The line y − x = 2 has slope m1 = 1 so we set m2 = dx √ sec x = ± 2 so x = nπ ± π/4 where n = 0, ±1, ±2, . . ..

lim (x2 − 1) = lim k(x − 1) = 0 = f (1), so any value of k gives continuity at x = 1. +

x→1−

x→1

lim f (x) = lim 2x = 2, and lim f (x) = lim k = k, so only if k = 2 is f (x) diﬀeren-

x→1−

x→1+

tiable at x = 1.

x→1+

15. The slope of the tangent line is the derivative y = 2x 1 = a + b. The slope of the secant is 2 2 x= 2 (a+b) a −b = a + b, so they are equal. a−b

(b, b 2)

(a, a 2) x a a+b b 2

16. To average 60 mi/h one would have to complete the trip in two hours. At 50 mi/h, 100 miles are completed after two hours. Thus time is up, and the speed for the remaining 20 miles would have to be inﬁnite. −1 −1 ∆x = (−0.5) = 0.5; and 2 (2 − 1)2 (x − 1)

17. (a) ∆x = 1.5 − 2 = −0.5; dy =

∆y =

√

25 − 32 −

√

−x −0 = (3) = 0; and 25 − x2 25 − (0)2

1 1 − = 2 − 1 = 1. (1.5 − 1) (2 − 1) (b) ∆x = 0 − (−π/4) = π/4; dy = sec2 (−π/4) (π/4) = π/2; and ∆y = tan 0 − tan(−π/4) = 1. ∆y =

25 − 02 = 4 − 5 = −1. (dV /d+)| =5 = 3+2 =5 = 3(5)2 = 75

18. (a)

43 − 23 56 = = 28 4−2 2

19. (a)

dW dW = 200(t − 15); at t = 5, = −2000; the water is running out at the rate of 2000 dt dt gal/min.

W (5) − W (0) 10000 − 22500 = = −2500; the average rate of ﬂow out is 2500 gal/min. 5−0 5

(b)

46π π 46π ; let x0 = and x = . Then 180 4 180

π 46π π π 2 π ◦ x− =1−2 = 0.9651; − cot 46 = cot x ≈ cot − csc 4 4 4 180 4 with a calculator, cot 46◦ = 0.9657.

uh a

20. cot 46◦ = cot

51 21. (a) h = 115 tan φ, dh = 115 sec2 φ dφ; with φ = 51◦ = π radians and 180 π

radians, h ± dh = 115(1.2349) ± 2.5340 = 142.0135 ± 2.5340, so dφ = ±0.5◦ = ±0.5 180 the height lies between 139.48 m and 144.55 m.

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Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Chapter 3 Supplementary Exercises

51 5 cos2 π ≈ 0.017 radians, or |dφ| ≤ 0.98◦ . 115 180

1 dT 2 1 =√ √ =√ dL g2 L gL

(b) s/m

us uf

22. (a)

(b) If |dh| ≤ 5 then |dφ| ≤

119

dT (c) Since > 0 an increase in L gives an increase in T , which is the period. To speed up a dL clock, decrease the period; to decrease T , decrease L. √ L dT = − 3/2 < 0; a decrease in g will increase T and the clock runs slower (d) dg g √ √ dT L −1 −3/2 g (e) = − 3/2 (f ) s3 /m =2 L 2 dg g

23. (a) f (x) = 2x, f (1.8) = 3.6 (b) f (x) = (x2 − 4x)/(x − 2)2 , f (3.5) ≈ −0.777778

25. f (2) ≈ 2.772589; f (2) = 4 ln 2 26. f (2) = 0.312141; f (2) = 2sin 2 (cos 2 ln 2 + sin 2/2)

1 d 2.5 3(h + 1)2.5 + 580h − 3 1 = 58 + 3x = lim = 58 + (2.5)(3)(1)1.5 = 58.75 ft/s h→0 10h 10 dx 10 x=1

27. vinst

24. (a) f (x) = 3x2 − 2x, f (2.3) = 11.27 (b) f (x) = (1 − x2 )/(x2 + 1)2 , f (−0.5) = 0.48

28. 164 ft/s

2500

29. Solve 3x2 − cos x = 0 to get x = ±0.535428.

30. When x4 − x − 1 > 0, f (x) = x4 − 2x − 1; when x4 − x − 1 < 0, f (x) = −x4 + 1, and f is diﬀerentiable in both cases. The roots of x4 − x − 1 = 0 are x1 = −0.724492, x2 = 1.220744. So x4 − x − 1 > 0 on (−∞, x1 ) and (x2 , +∞), and x4 −x−1 < 0 on (x1 , x2 ). Then lim f (x) = x→x− 1

uh a

-1.5

lim (4x3 − 2) = 4x31 − 2 and

x→x− 1 3

lim f (x) = lim+ −4x = −4x31 which is not equal

x→x+ 1

1.5

x→x1

-1.5

− 2, so f is not diﬀerentiable at x = x1 ; simito larly f is not diﬀerentiable at x = x2 . 4x31

31. (a) f (x) = 5x4 (d) f (x) = −3/(x − 1)2

(b) f (x) = −1/x2 √ (e) f (x) = 3x/ 3x2 + 5

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32. f (x) = 2x sin x + x2 cos x

36. f (x) =

37. f (x) =

6x2 + 8x − 17 (3x + 2)2 x2 cos

√

√ 1 − 2 x sin 2x √ 2 x

35. f (x) =

(1 + x2 ) sec2 x − 2x tan x (1 + x2 )2

√ x − 2x3/2 sin x 2x7/2

−2x5 sin x − 2x4 cos x + 4x4 + 6x2 sin x + 6x − 3x cos x − 4x sin x + 4 cos x − 8 √ 2x2 x4 − 3 + 2(2 − cos x)2

34. f (x) =

33. f (x) =

Chapter 3

us uf

120

2 −1/3 2 −1/3 − y y − y = 0. At x = 1 and y = −1, y = 2. The tangent line is x 3 3 y + 1 = 2(x − 1).

38. Diﬀerentiating,

dθ dy dz = a and = −b. From the given dt x=1 dt dt y=1 √ ﬁgure sin θ = y/z; when x = y = 1, z = 2. So

40. Find

39. Diﬀerentiating, (xy + y) cos xy = y . With x = π/2 and y = 1 this becomes y = 0, so the equation of the tangent line is y − 1 = 0(x − π/2) or y = 1.

when x = y = 1.

θ = sin−1 (y/z) and y dz a 1 dθ 1 dy − 2 = −b − √ = z dt dt 2 1 − y 2 /z 2 z dt

CHAPTER 3 HORIZON MODULE

1. x1 = l1 cos θ1 , x2 = l2 cos(θ1 + θ2 ), so x = x1 + x2 = l1 cos θ1 + l2 cos(θ1 + θ2 ) (see Figure 3 in text); similarly y1 = l1 sin θ1 + l2 sin(θ1 + θ2 ).

2. Fix θ1 for the moment and let θ2 vary; then the distance r from (x, y) to the origin (see Figure 3 in text) is at most l1 + l2 and at least l1 − l2 if l1 ≥ l2 and l2 − l1 otherwise. For any ﬁxed θ2 let θ1 vary and the point traces out a circle of radius r.

(x, y) = (l1 cos θ + l2 cos(θ1 + θ2 ), l1 sin θ1 + l2 sin(θ1 + θ2 ))

uh a

(a) {(x, y) : 0 ≤ x2 + y 2 ≤ 2l1 } (b) {(x, y) : l1 − l2 ≤ x2 + y 2 ≤ l1 + l2 } (c) {(x, y) : l2 − l1 ≤ x2 + y 2 ≤ l1 + l2 }

= (cos(π/4) + 3 cos(5π/12), sin(π/4) + 3 sin(5π/12)) =

√

√ √ √ 2+3 6 7 2+3 6 , 4 4

4. x = (1) cos 2t + (1) cos(2t + 3t) = cos 2t + cos 5t, y = (1) sin 2t + (1) sin(2t + 3t) = sin 2t + sin 5t

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Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Chapter 3 Horizon Module

121

-2

-2 v1 = 1, v2 = 4

-2 v1 = 3, v2 = 5

-2 v1 = 4, v2 = 1

6. x = 2 cos t, y = 2 sin t, a circle of radius 2

x 2

us uf

x -2

9 = [3 sin(θ1 + θ2 )]2 + [3 cos(θ1 + θ2 )]2 = [5 − 3 sin θ1 ]2 + [3 − 3 cos θ1 ]2 = 25 − 30 sin θ1 + 9 sin2 θ1 + 9 − 18 cos θ1 + 9 cos2 θ1 = 43 − 30 sin θ1 − 18 cos θ1 , so 15 sin θ1 + 9 cos θ1 = 17 2 17 − 9 cos θ1 2 2 (b) 1 = sin θ1 + cos θ2 = + cos θ1 , or 306 cos2 θ1 − 306 cos θ1 = −64 15 √

1 5 17 2 (c) cos θ1 = 153 ± (153) − 4(153)(32) /306 = ± 2 102 (e) If θ1 = 0.792436 rad, then θ2 = 0.475882 rad ≈ 27.2660◦ ; if θ1 = 1.26832 rad, then θ2 = −0.475882 rad ≈ −27.2660◦ .

dθ1 dθ2 + dt dt

dθ1 dx = −3 sin θ1 − (3 sin(θ1 + θ2 )) dt dt

dθ1 dθ2 (sin θ1 + sin(θ1 + θ2 )) − 3 (sin(θ1 + θ2 )) dt dt

= −y

dθ2 dθ1 − 3(sin(θ1 + θ2 )) ; dt dt

similarly

= −3

dθ2 dy dθ1 dx dy =x + 3(cos(θ1 + θ2 )) . Now set = 0, = 1. dt dt dt dt dt

7. (a)

uh a

1 9. (a) x = 3 cos(π/3) + 3 cos(−π/3) = 6 = 3 and y = 3 sin(π/3) − 3 sin(π/3) = 0; equations (4) 2 dθ2 dθ1 dθ2 become 3 sin(π/3) = 0, 3 +3 cos(π/3) = 1 with solution dθ2 /dt = 0, dθ1 /dt = 1/3. dt dt dt dθ1 dθ2 dθ1 = 0 and −3 −3 = 1, with solution dθ1 /dt = 0, (b) x = −3, y = 3, so −3 dt dt dt dθ2 /dt = −1/3.

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Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH CHAPTER 4

us uf

Exponential, Logarithmic, and Inverse Trigonometric Functions EXERCISE SET 4.1

f (g(x)) = 4(x/4) = x, g(f (x)) = (4x)/4 = x, f and g are inverse functions f (g(x)) = 3(3x − 1) + 1 = 9x − 2 = x so f and g are not inverse functions f (g(x)) = 3 (x3 + 2) − 2 = x, g(f (x)) = (x − 2) + 2 = x, f and g are inverse functions f (g(x)) = (x1/4 )4 = x, g(f (x)) = (x4 )1/4 = |x| = x, f and g are not inverse functions

1. (a) (b) (c) (d)

2. (a) They are inverse functions.

-2 2

-2

(b) The graphs are not reﬂections of each other about the line y = x.

-2

5 0

(d) They are inverse functions provided the domain of f (x) is restricted to [0, +∞)

uh a

2 0

3. (a) yes; all outputs (the elements of row two) are distinct (b) no; f (1) = f (6)

122

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Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Exercise Set 4.1

123

(b) yes

(d) yes

6. (a) no, the horizontal line test fails

(e) no

(b) yes, horizontal line test

-1 -3

3 -10

-2

(f ) no

5. (a) yes

us uf

4. (a) no; it is easy to conceive of, say, 8 people in line at two diﬀerent times (b) no; perhaps your weight remains constant for more than a year (c) yes, since the function is increasing, in the sense that the greater the volume, the greater the weight

8. (d) no, the horizontal line test fails (e) no, the horizontal line test fails (f ) yes, horizontal line test

7. (a) no, the horizontal line test fails (b) no, the horizontal line test fails (c) yes, horizontal line test

9. (a) f has an inverse because the graph passes the horizontal line test. To compute f −1 (2) start at 2 on the y-axis and go to the curve and then down, so f −1 (2) = 8; similarly, f −1 (−1) = −1 and f −1 (0) = 0.

(b) domain of f −1 is [−2, 2], range is [−8, 8]

(c)

8 4 x -2

-4 -8

10. (a) the horizontal line test fails (b) −∞ < x ≤ −1; −1 ≤ x ≤ 2; and 2 ≤ x < 4.

11. (a) f (x) = 2x + 8; f < 0 on (−∞, −4) and f > 0 on (−4, +∞); not one-to-one (b) f (x) = 10x4 + 3x2 + 3 ≥ 3 > 0; f (x) is positive for all x, so f is one-to-one (c) f (x) = 2 + cos x ≥ 1 > 0 for all x, so f is one-to-one

uh a

12. (a) f (x) = 3x2 + 6x = x(3x + 6) changes sign at x = −2, 0, so f is not one-to-one (b) f (x) = 5x4 + 24x2 + 2 ≥ 2 > 0; f is positive for all x, so f is one-to-one 1 ; f is one-to-one because: (c) f (x) = (x + 1)2 if x1 < x2 < −1 then f > 0 on [x1 , x2 ], so f (x1 ) = f (x2 ) if −1 < x1 < x2 then f > 0 on [x1 , x2 ], so f (x1 ) = f (x2 ) if x1 < −1 < x2 then f (x1 ) > 1 > f (x2 ) since f (x) > 1 on (−∞, −1) and f (x) < 1 on (−1, +∞)

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124

Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Chapter 4

13. y = f −1 (x), x = f (y) = y 5 , y = x1/5 = f −1 (x)

15. y = f −1 (x), x = f (y) = 7y − 6, y =

1 x = f −1 (x) 6

us uf

14. y = f −1 (x), x = f (y) = 6y, y =

1 (x + 6) = f −1 (x) 7

y+1 x+1 = f −1 (x) , xy − x = y + 1, (x − 1)y = x + 1, y = y−1 x−1 17. y = f −1 (x), x = f (y) = 3y 3 − 5, y = 3 (x + 5)/3 = f −1 (x)

18. y = f −1 (x), x = f (y) =

√ 5

4y + 2, y =

√ 3

1 5 (x − 2) = f −1 (x) 4

2y − 1, y = (x3 + 1)/2 = f −1 (x) 5−x 5 −1 20. y = f (x), x = f (y) = 2 ,y= = f −1 (x) x y +1 21. y = f −1 (x), x = f (y) = 3/y 2 , y = − 3/x = f −1 (x) 2y, y ≤ 0 x/2, x ≤ 0 −1 −1 , y = f (x) = 22. y = f (x), x = f (y) = √ 2 x, x > 0 y , y>0 23. y = f

(x), x = f (y) =

5/2 − y,

y<2

1/y,

y≥2

, y=f

−1

(x) =

−1

19. y = f −1 (x), x = f (y) =

16. y = f −1 (x), x = f (y) =

5/2 − x,

x > 1/2

1/x, 0 < x ≤ 1/2

24. y = p−1 (x), x = p(y) = y 3 − 3y 2 + 3y − 1 = (y − 1)3 , y = x1/3 + 1 = p−1 (x)

25. y = f −1 (x), x = f (y) = (y + 2)4 for y ≥ 0, y = f −1 (x) = x1/4 − 2 for x ≥ 16 √ 26. y = f −1 (x), x = f (y) = y + 3 for y ≥ −3, y = f −1 (x) = x2 − 3 for x ≥ 0 √ 27. y = f −1 (x), x = f (y) = − 3 − 2y for y ≤ 3/2, y = f −1 (x) = (3 − x2 )/2 for x ≤ 0 28. y = f −1 (x), x = f (y) = 3y 2 + 5y − 2 for y ≥ 0, 3y 2 + 5y − 2 − x = 0 for y ≥ 0, √ y = f −1 (x) = (−5 + 12x + 49)/6 for x ≥ −2

29. y = f −1 (x), x = f (y) = y − 5y 2 for y ≥ 1, 5y 2 − y + x = 0 for y ≥ 1, √ y = f −1 (x) = (1 + 1 − 20x)/10 for x ≤ −4

5 (F − 32) 9 (b) how many degrees Celsius given the Fahrenheit temperature (c) C = −273.15◦ C is equivalent to F = −459.67◦ F, so the domain is F ≥ −459.67, the range is C ≥ −273.15

30. (a) C =

uh a

31. (a) y = f (x) = (6.214 × 10−4 )x (c) how many meters in y miles

(b) x = f −1 (y) =

104 y 6.214

32. f and f −1 are continuous so f (3) = lim f (x) = 7; then f −1 (7) = 3, and x→3 −1 −1 −1 lim x = f (7) = 3 lim f (x) = f x→7

x→7

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Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Exercise Set 4.1

√ 33. (a) f (g(x)) = f ( x) √ = ( x)2 = x, x > 1; g(f (x)) = g(x2 ) √ = x2 = x, x > 1

(b)

125

y = g(x)

us uf

y = f (x)

34. y = f −1 (x), x = f (y) = ay 2 + by + c, ay 2 + by + c − x = 0, use the quadratic formula to get −b ± b2 − 4a(c − x) y= ; 2a −b + b2 − 4a(c − x) −b − b2 − 4a(c − x) −1 −1 (b) f (x) = (a) f (x) = 2a 2a 3−x 1 − x = 3 − 3x − 3 + x = x so f = f −1 35. (a) f (f (x)) = 3−x 1−x−3+x 1− 1−x

3−

(b) symmetric about the line y = x

36. y = m(x − x0 ) is an equation of the line. The graph of the inverse of f (x) = m(x − x0 ) will be the reﬂection of this line about y = x. Solve y = m(x − x0 ) for x to get x = y/m + x0 = f −1 (y) so y = f −1 (x) = x/m + x0 .

37. (a) f (x) = x3 − 3x2 + 2x = x(x − 1)(x − 2) so f (0) = f (1) = f (2) = 0 thus f is not one-to-one. √ √ 6 ± 36 − 24 = 1 ± 3/3. f (x) > 0 (f is (b) f (x) = 3x2 − 6x + 2, f (x) = 0 when x = 6 √ √ √ increasing) if x < 1 − 3/3, f (x) √ < 0 (f is decreasing) if 1√− 3/3 < x <√1 + 3/3, so f (x) takes on values less than f (1 − 3/3) on both sides of 1 − 3/3 thus 1 − 3/3 is the largest value of k.

38. (a) f (x) = x3 (x − 2) so f (0) = f (2) = 0 thus f is not one to one. (b) f (x) = 4x3 − 6x2 = 4x2 (x − 3/2), f (x) = 0 when x = 0 or 3/2; f is decreasing on (−∞, 3/2] and increasing on [3/2, +∞) so 3/2 is the smallest value of k.

39. if f −1 (x) = 1, then x = f (1) = 2(1)3 + 5(1) + 3 = 10

40. if f −1 (x) = 2, then x = f (2) = (2)3 /[(2)2 + 1] = 8/5 41.

uh a -2

42.

-2

-5

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Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

126

Chapter 4

43.

44.

dx dy 1 = 15y 2 + 1, = ; dy dx 15y 2 + 1

45. y = f −1 (x), x = f (y) = 5y 3 + y − 7,

dy dy 1 dy + , = 15y 2 + 1 dx dx dx

46. y = f −1 (x), x = f (y) = 1/y 2 , check: 1 = −2y −3

dx dy = −2y −3 , = −y 3 /2; dy dx

dy dy , = −y 3 /2 dx dx

check: 1 = 15y 2

us uf

dx dy 1 ; = 10y 4 + 3y 2 , = 4 dy dx 10y + 3y 2 dy dy dy 1 + 3y 2 , check: 1 = 10y 4 = dx dx 10y 4 + 3y 2 dx

47. y = f −1 (x), x = f (y) = 2y 5 + y 3 + 1,

dy 1 dx = 5 − 2 cos 2y, = ; dy dx 5 − 2 cos 2y dy dy 1 check: 1 = (5 − 2 cos 2y) , = dx dx 5 − 2 cos 2y

48. y = f −1 (x), x = f (y) = 5y − sin 2y,

49. f (f (x)) = x thus f = f −1 so the graph is symmetric about y = x.

51.

50. (a) Suppose x1 = x2 where x1 and x2 are in the domain of g and g(x1 ), g(x2 ) are in the domain of f then g(x1 ) = g(x2 ) because g is one-to-one so f (g(x1 )) = f (g(x2 )) because f is one-to-one thus f ◦ g is one-to-one because (f ◦ g)(x1 ) = (f ◦ g)(x2 ) if x1 = x2 . (b) f , g, and f ◦ g all have inverses because they are all one-to-one. Let h = (f ◦ g)−1 then (f ◦ g)(h(x)) = f [g(h(x))] = x, apply f −1 to both sides to get g(h(x)) = f −1 (x), then apply g −1 to get h(x) = g −1 (f −1 (x)) = (g −1 ◦ f −1 )(x), so h = g −1 ◦ f −1

uh a

52. Suppose that g and h are both inverses of f then f (g(x)) = x, h[f (g(x))] = h(x), but h[f (g(x))] = g(x) because h is an inverse of f so g(x) = h(x).

53. F (x) = 2f (2g(x))g (x) so F (3) = 2f (2g(3))g (3). By inspection f (1) = 3, so g(3) = f −1 (3) = 1 and g (3) = (f −1 ) (3) = 1/f (f −1 (3)) = 1/f (1) = 1/7 because f (x) = 4x3 + 3x2 . Thus F (3) = 2f (2)(1/7) = 2(44)(1/7) = 88/7. F (3) = f (2g(3)), g(3) = f −1 (3); by inspection f (1) = 3, so g(3) = f −1 (3) = 1, F (3) = f (2) = 25.

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Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Exercise Set 4.2

127

(b) 4

2. (a) 1/16

(b) 8

3. (a) 2.9690

(b) 0.0341

4. (a) 1.8882

(b) 0.9381

1 = log2 (2−5 ) = −5 (b) log2 32 (d) log9 3 = log9 (91/2 ) = 1/2

5. (a) log2 16 = log2 (2 ) = 4 (c) log4 4 = 1 6. (a) log10 (0.001) = log10 (10−3 ) = −3

log10 (104 ) = 4 √ ln( e) = ln(e1/2 ) = 1/2

(b)

(d)

(b) −0.3011

8. (a) −0.5229

(b) 1.1447

1 1 ln b + ln c = 2r + s/2 + t/2 2 2

24 (16) 13. log = log(256/3) 3 15. ln √

(b)

1 (ln a + 3 ln b − 2 ln c) = r/2 + 3s/2 − t 2

(b) 2 ln |x| + 3 ln sin x −

(b)

14. log

√

1 ln(x2 + 1) 2

1 1 ln(x2 + 1) − ln(x3 + 5) 2 2

√ 100 x x − log(sin 2x) + log 100 = log sin3 2x 3

16. 1 + x = 103 = 1000, x = 999

x = 10−1 = 0.1, x = 0.01

17.

x(x + 1)2 cos x

√ 3

1 log(x + 2) − log cos 5x 3

1 log(x − 3) 2

(b) ln b − 3 ln a − ln c = s − 3r − t

1 ln c − ln a − ln b = t/3 − r − s 3

11. (a) 1 + log x + 12. (a)

7. (a) 1.3655

10. (a)

9. (a) 2 ln a +

us uf

1. (a) −4

EXERCISE SET 4.2

19. 1/x = e−2 , x = e2

20. x = 7 22. log10 x3 = 30, x3 = 1030 , x = 1010

uh a

21. 2x = 8, x = 4

18. x2 = e4 , x = ±e2

23. log10 x = 5, x = 105 √ 4 4 = ln 2, 5 = 2, x5 = 2, x = 5 2 5 x x 25. ln 2x2 = ln 3, 2x2 = 3, x2 = 3/2, x = 3/2 (we discard − 3/2 because it does not satisfy the original equation)

24. ln 4x − ln x6 = ln 2, ln

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Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Chapter 4

ln 2 ln 3

27. ln 5−2x = ln 3, −2x ln 5 = ln 3, x = −

26. ln 3x = ln 2, x ln 3 = ln 2, x =

ln 3 2 ln 5

us uf

128

1 28. e−2x = 5/3, −2x = ln(5/3), x = − ln(5/3) 2 1 ln(7/2) 3

29. e3x = 7/2, 3x = ln(7/2), x =

31. e−x (x + 2) = 0 so e−x = 0 (impossible) or x + 2 = 0, x = −2

30. ex (1 − 2x) = 0 so ex = 0 (impossible) or 1 − 2x = 0, x = 1/2

32. e2x − ex − 6 = (ex − 3)(ex + 2) = 0 so ex = −2 (impossible) or ex = 3, x = ln 3 33. e−2x − 3e−x + 2 = (e−x − 2)(e−x − 1) = 0 so e−x = 2, x = − ln 2 or e−x = 1, x = 0 y

(b)

2 x 6

34. (a)

-2

35. (a) 6

2 x -2

2 y

(b)

2 x

-4

36. (a)

-2

-4

(b)

uh a

-1

-10

x 3 -1

37. log2 7.35 = (log 7.35)/(log 2) = (ln 7.35)/(ln 2) ≈ 2.8777; log5 0.6 = (log 0.6)/(log 5) = (ln 0.6)/(ln 5) ≈ −0.3174

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Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Exercise Set 4.2

39.

us uf

0 2

-5

38.

129

-3

40. (a) Let X = logb x and Y = loga x. Then bX = x and aY = x so aY = bX , or aY /X = b, which loga x loga x means loga b = Y /X. Substituting for Y and X yields = loga b, logb x = . logb x loga b (b) Let x = a to get logb a = (loga a)/(loga b) = 1/(loga b) so (loga b)(logb a) = 1. (log2 81)(log3 32) = (log2 [34 ])(log3 [25 ]) = (4 log2 3)(5 log3 2) = 20(log2 3)(log3 2) = 20 (b) x ≈ 332105.11, y ≈ 12.7132

41. (a) x = 3.6541, y = 1.2958

12.7134

332085 12.7130

332125

2 0.6

42. Since the units are billions, one trillion is 1,000 units. Solve 1000 = 0.051517(1.1306727)x for x by taking common logarithms, resulting in 3 = log 0.051517 + x log 1.1306727, which yields x ≈ 77.4, so the debt ﬁrst reached one trillion dollars around 1977.

43. (a) no, the curve passes through the origin −x

y = 2x/4 √ (d) y = ( 5)x

(b)

-1

2 0

44. (a) As x → +∞ the function grows very slowly, but it is always increasing and tends to +∞. As x → 1+ the function tends to −∞. y

(b)

uh a

x 1

-5

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130

Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Chapter 4

47. 75e−t/125 = 15, t = −125 ln(1/5) = 125 ln 5 ≈ 201 days.

50. (a) log[H + ] = −2.44, [H + ] = 10−2.44 ≈ 3.6 × 10−3 mol/L (b) log[H + ] = −8.06, [H + ] = 10−8.06 ≈ 8.7 × 10−9 mol/L

(b) 120 dB; damage (d) 75 dB; no damage

51. (a) 140 dB; damage (c) 80 dB; no damage

(d) 5.9; acidic

(b) 4.2; acidic

(b) Q = 12e−0.055(4) = 12e−0.22 ≈ 9.63 grams

48. (a) If t = 0, then Q = 12 grams

49. (a) 7.4; basic

us uf

46. Let x = logb a and y = logb c, so a = bx and c = by . First, ac = bx by = bx+y or equivalently, logb (ac) = x + y = logb a + logb c. Secondly, a/c = bx /by = bx−y or equivalently, logb (a/c) = x − y = logb a − logb c. Next, ar = (bx )r = brx or equivalently, logb ar = rx = r logb a. Finally, 1/c = 1/by = b−y or equivalently, logb (1/c) = −y = − logb c.

45. log(1/2) < 0 so 3 log(1/2) < 2 log(1/2)

52. Suppose that I1 = 3I2 and β1 = 10 log10 I1 /I0 , β2 = 10 log10 I2 /I0 . Then I1 /I0 = 3I2 /I0 , log10 I1 /I0 = log10 3I2 /I0 = log10 3 + log10 I2 /I0 , β1 = 10 log10 3 + β2 , β1 − β2 = 10 log10 3 ≈ 4.8 decibels. 53. Let IA and IB be the intensities of the automobile and blender, respectively. Then log10 IA /I0 = 7 and log10 IB /I0 = 9.3, IA = 107 I0 and IB = 109.3 I0 , so IB /IA = 102.3 ≈ 200.

54. The decibel level of the nth echo is 120(2/3)n ; log(1/12) log 12 120(2/3)n < 10 if (2/3)n < 1/12, n < = ≈ 6.13 so 6 echoes can be heard. log(2/3) log 1.5

55. (a) log E = 4.4 + 1.5(8.2) = 16.7, E = 1016.7 ≈ 5 × 1016 J (b) Let M1 and M2 be the magnitudes of earthquakes with energies of E and 10E, respectively. Then 1.5(M2 − M1 ) = log(10E) − log E = log 10 = 1, M2 − M1 = 1/1.5 = 2/3 ≈ 0.67. 56. Let E1 and E2 be the energies of earthquakes with magnitudes M and M + 1, respectively. Then log E2 − log E1 = log(E2 /E1 ) = 1.5, E2 /E1 = 101.5 ≈ 31.6.

57. If t = −2x, then x = −t/2 and lim (1 − 2x)1/x = lim (1 + t)−2/t = lim [(1 + t)1/t ]−2 = e−2 . x→0

t→0

58. If t = 3/x, then x = 3/t and lim (1 + 3/x)x = lim+ (1 + t)3/t = lim+ [(1 + t)1/t ]3 = e3 . x→+∞

t→0

uh a

EXERCISE SET 4.3 1.

1 (2) = 1/x 2x

2 ln x 1 = 3. 2(ln x) x x

1 (3x2 ) = 3/x x3

1 (cos x) = cot x sin x

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Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Exercise Set 4.3

(1 + x2 )(1) − x(2x) 1 1 − x2 = (1 + x2 )2 x(1 + x2 ) x/(1 + x2 )

1 ln x

1 1 = x x ln x

1 10. x + (3x2 ) ln x = x2 (1 + 3 ln x) x 3

11.

1 2 2(ln x)(1/x) 2

ln x = 1 + ln x x 1 + ln2 x

2 x

1 √ = √ 2 x(2 + x)

3x2 − 14x − 7x2 − 3

1 (ln x)−1/2 2

1 1 = √ x 2x ln x

15. 3x2 log2 (3 − 2x) +

−2x3 (ln 2)(3 − 2x)

sin(2 ln x) sin(ln x2 ) 1 = = x x x

2 log2 (x2 − 2x) + 3x log2 (x2 − 2x)

17.

2x(1 + log x) − x/(ln 10) (1 + log x)2

(x2

2x − 2 − 2x) ln 2

16.

18. 1/[x(ln 10)(1 + log x)2 ]

21. x3 ex + 3x2 ex = x2 ex (x + 3)

20. −10xe−5x

22. −

1 1/x e x2

(ex + e−x )(ex + e−x ) − (ex − e−x )(ex − e−x ) dy = dx (ex + e−x )2 (e2x + 2 + e−2x ) − (e2x − 2 + e−2x ) = 4/(ex + e−x )2 (ex + e−x )2

23.

1 √

1 13. − sin(ln x) x

14. 2 sin(ln x) cos(ln x)

19. 7e7x

12.

1 √ 2+ x

us uf

1 sec2 x (sec2 x) = tan x tan x

24. ex cos(ex )

uh a

25. (x sec2 x + tan x)ex tan x

27. (1 − 3e3x )e(x−e

29.

)

(x − 1)e−x x−1 = x 1 − xe−x e −x

131

26.

(ln x)ex − ex (1/x) dy ex (x ln x − 1) = = dx (ln x)2 x(ln x)2

28.

15 2 x (1 + 5x3 )−1/2 exp( 1 + 5x3 ) 2

30.

1 [− sin(ex )]ex = −ex tan(ex ) cos(ex )

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Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Chapter 4

32.

1 dy = dx x tan y

33.

1 3x d ln cos x − ln(4 − 3x2 ) = − tan x + dx 2 4 − 3x2

34.

d dx

= 0,

dy y =− dx x(y + 1)

dy tan y dy x sec y + tan y , = dx dx x(tan y − sec2 y)

dy +y dx

1 dy + dx xy

us uf

31.

1 1 1 1 [ln(x − 1) − ln(x + 1)] = − 2 2 x−1 x+1

1 2x 1 dy 3 2 2 35. ln |y| = ln |x| + ln |1 + x |, + =x 1+x x 3(1 + x2 ) 3 dx

1 1 x−1 − x+1 x−1 x+1

1 dy 1 36. ln |y| = [ln |x − 1| − ln |x + 1|], = 5 5 dx

132

1 1 ln |x2 − 8| + ln |x3 + 1| − ln |x6 − 7x + 5| 2 3 √ 3x2 6x5 − 7 dy (x2 − 8)1/3 x3 + 1 2x + − = 3(x2 − 8) 2(x3 + 1) x6 − 7x + 5 dx x6 − 7x + 5

37. ln |y| =

1 ln |x| 2

38. ln |y| = ln | sin x| + ln | cos x| + 3 ln | tan x| −

sin x cos x tan3 x 1 dy 3 sec2 x √ = − cot x − tan x + dx tan x 2x x 39. f (x) = 2x ln 2; y = 2x , ln y = x ln 2,

1 y = ln 2, y = y ln 2 = 2x ln 2 y

40. f (x) = −3−x ln 3; y = 3−x , ln y = −x ln 3,

1 y = − ln 3, y = −y ln 3 = −3−x ln 3 y

41. f (x) = π sin x (ln π) cos x;

y = π sin x , ln y = (sin x) ln π,

1 y = (ln π) cos x, y = π sin x (ln π) cos x y

42. f (x) = π x tan x (ln π)(x sec2 x + tan x);

uh a

y = π x tan x , ln y = (x tan x) ln π,

1 y = (ln π)(x sec2 x + tan x) y

y = π x tan x (ln π)(x sec2 x + tan x)

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Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Exercise Set 4.3

1 dy 3x2 − 2 1 = 3 ln x + ln(x3 − 2x), y dx x − 2x x 2 3x − 2 dy 1 = (x3 − 2x)ln x 3 ln x + ln(x3 − 2x) x − 2x x dx

133

1 dy sin x dy sin x sin x 44. ln y = (sin x) ln x, = + (cos x) ln x, =x + (cos x) ln x y dx x dx x 1 dy 1 = tan x + (sec2 x) ln(ln x), y dx x ln x dy tan x = (ln x)tan x + (sec2 x) ln(ln x) x ln x dx

2x 1 1 dy = 2 ln x + ln(x2 + 3), y dx x +3 x dy 2x 1 2 ln x 2 = (x + 3) ln x + ln(x + 3) dx x2 + 3 x

46. ln y = (ln x) ln(x2 + 3),

45. ln y = (tan x) ln(ln x),

us uf

43. ln y = (ln x) ln(x3 − 2x),

47. f (x) = exe−1

(b) logx 2 =

1 d ln e 1 = , [logx e] = − ln x ln x dx x(ln x)2 ln 2 d ln 2 , [logx 2] = − ln x dx x(ln x)2

49. (a) logx e =

48. (a) because xx is not of the form ax where a is constant 1 (b) y = xx , ln y = x ln x, y = 1 + ln x, y = xx (1 + ln x) y

ln b ln e 1 50. (a) From loga b = for a, b > 0 it follows that log(1/x) e = =− , hence ln a ln(1/x) ln x

1 d log(1/x) e = dx x(ln x)2 1 ln e 1 1 d 1 log(ln x) e = − (b) log(ln x) e = = , so =− ln(ln x) ln(ln x) dx (ln(ln x))2 x ln x x(ln x)(ln(ln x))2

dy = e−λt (ωA cos ωt − ωB sin ωt) + (−λ)e−λt (A sin ωt + B cos ωt) dt

uh a

52.

51. (a) f (x) = kekx , f (x) = k 2 ekx , f (x) = k 3 ekx , . . . , f (n) (x) = k n ekx (b) f (x) = −ke−kx , f (x) = k 2 e−kx , f (x) = −k 3 e−kx , . . . , f (n) (x) = (−1)n k n e−kx

= e−λt [(ωA − λB) cos ωt − (ωB + λA) sin ωt]

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Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

134

Chapter 4

2 2 1 xâ&#x2C6;&#x2019;Âľ d 1 1 xâ&#x2C6;&#x2019;Âľ â&#x2C6;&#x2019; 53. f (x) = â&#x2C6;&#x161; exp â&#x2C6;&#x2019; 2 Ď&#x192; dx 2 Ď&#x192; 2Ď&#x20AC;Ď&#x192; 2 1 1 xâ&#x2C6;&#x2019;Âľ 1 xâ&#x2C6;&#x2019;Âľ â&#x2C6;&#x161; = â&#x2C6;&#x2019; exp â&#x2C6;&#x2019; 2 Ď&#x192; Ď&#x192; Ď&#x192; 2Ď&#x20AC;Ď&#x192; 2 1 1 xâ&#x2C6;&#x2019;Âľ = â&#x2C6;&#x2019;â&#x2C6;&#x161; (x â&#x2C6;&#x2019; Âľ) exp â&#x2C6;&#x2019; 2 Ď&#x192; 2Ď&#x20AC;Ď&#x192; 3

us uf

54. y = Aekt , dy/dt = kAekt = k(Aekt ) = ky

55. y = Ae2x + Beâ&#x2C6;&#x2019;4x , y = 2Ae2x â&#x2C6;&#x2019; 4Beâ&#x2C6;&#x2019;4x , y = 4Ae2x + 16Beâ&#x2C6;&#x2019;4x so y + 2y â&#x2C6;&#x2019; 8y = (4Ae2x + 16Beâ&#x2C6;&#x2019;4x ) + 2(2Ae2x â&#x2C6;&#x2019; 4Beâ&#x2C6;&#x2019;4x ) â&#x2C6;&#x2019; 8(Ae2x + Beâ&#x2C6;&#x2019;4x ) = 0

(b) y = â&#x2C6;&#x2019;x2 eâ&#x2C6;&#x2019;x

+ eâ&#x2C6;&#x2019;x

= eâ&#x2C6;&#x2019;x

(1 â&#x2C6;&#x2019; x2 ), xy = xeâ&#x2C6;&#x2019;x

56. (a) y = â&#x2C6;&#x2019;xeâ&#x2C6;&#x2019;x + eâ&#x2C6;&#x2019;x = eâ&#x2C6;&#x2019;x (1 â&#x2C6;&#x2019; x), xy = xeâ&#x2C6;&#x2019;x (1 â&#x2C6;&#x2019; x) = y(1 â&#x2C6;&#x2019; x) /2

(1 â&#x2C6;&#x2019; x2 ) = y(1 â&#x2C6;&#x2019; x2 )

1 ln(1 + h) â&#x2C6;&#x2019; ln 1 ln(1 + h) = lim = 57. (a) f (w) = ln w; f (1) = lim =1 hâ&#x2020;&#x2019;0 hâ&#x2020;&#x2019;0 h h w w=1 d 10h â&#x2C6;&#x2019; 1 w w w = (10 ) (b) f (w) = 10 ; f (0) = lim = 10 ln 10 = ln 10 hâ&#x2020;&#x2019;0 h dw w=0 w=0

d ln(e2 + â&#x2C6;&#x2020;x) â&#x2C6;&#x2019; 2 1 = (ln x) 58. (a) f (x) = ln x; f (e ) = lim = = eâ&#x2C6;&#x2019;2 â&#x2C6;&#x2020;xâ&#x2020;&#x2019;0 â&#x2C6;&#x2020;x dx x x=e2 x=e2 2w â&#x2C6;&#x2019; 2 d w w w (b) f (w) = 2 ; f (1) = lim = 2 ln 2 = 2 ln 2 = (2 ) wâ&#x2020;&#x2019;1 w â&#x2C6;&#x2019; 1 dw w=1 w=1 2

EXERCISE SET 4.4 1. (a) â&#x2C6;&#x2019;Ď&#x20AC;/2

(b) Ď&#x20AC;

2. (a) Ď&#x20AC;/3

(b) Ď&#x20AC;/3

(d) 0

(d) 2Ď&#x20AC;/3

â&#x2C6;&#x161; â&#x2C6;&#x161; â&#x2C6;&#x161; 3. Î¸ = â&#x2C6;&#x2019;Ď&#x20AC;/3; cos Î¸ = 1/2, tan Î¸ = â&#x2C6;&#x2019; 3, cot Î¸ = â&#x2C6;&#x2019;1/ 3, sec Î¸ = 2, csc Î¸ = â&#x2C6;&#x2019;2/ 3 â&#x2C6;&#x161;

4. Î¸ = Ď&#x20AC;/3; sin Î¸ =

3/2, tan Î¸ =

â&#x2C6;&#x161;

â&#x2C6;&#x161; â&#x2C6;&#x161; 3, cot Î¸ = 1/ 3, sec Î¸ = 2, csc Î¸ = 2/ 3

â?Ş

uh a

5. tan Î¸ = 4/3, 0 < Î¸ < Ď&#x20AC;/2; use the triangle shown to get sin Î¸ = 4/5, cos Î¸ = 3/5, cot Î¸ = 3/4, sec Î¸ = 5/3, csc Î¸ = 5/4

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Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Exercise Set 4.4

2.6

2.4

6. sec θ = 2.6, 0 < θ < π/2; use the triangle shown to get sin θ = 2.4/2.6 = 12/13, cos θ = 1/2.6 = 5/13, tan θ = 2.4 = 12/5, cot θ = 5/12, csc θ = 13/12

135

us uf

␪ 1

π/7 sin−1 (sin π) = sin−1 (sin 0) = 0 sin−1 (sin(5π/7)) = sin−1 (sin(2π/7)) = 2π/7 Note that π/2 < 630 − 200π < π so sin(630) = sin(630 − 200π) = sin(π − (630 − 200π)) = sin(201π − 630) where 0 < 201π − 630 < π/2; sin−1 (sin 630) = sin−1 (sin(201π − 630)) = 201π − 630.

8. (a) (b) (c) (d)

π/7 π cos−1 (cos(12π/7)) = cos−1 (cos(2π/7)) = 2π/7 Note that −π/2 < 200 − 64π < 0 so cos(200) = cos(200 − 64π) = cos(64π − 200) where 0 < 64π − 200 < π/2; cos−1 (cos 200) = cos−1 (cos(64π − 200)) = 64π − 200.

7. (a) (b) (c) (d)

9. (a) 0 ≤ x ≤ π (c) −π/2 < x < π/2

(b) −1 ≤ x ≤ 1 (d) −∞ < x < +∞

10. Let θ = sin−1 (−3/4) then sin θ = −3/4, −π/2 < θ < 0 and √ (see ﬁgure) sec θ = 4/ 7 √7

␪

-3

11. Let θ = cos−1 (3/5), sin 2θ = 2 sin θ cos θ = 2(4/5)(3/5) = 24/25 5 ␪

ad √

12. (a) sin(cos

−1

uh a

cos-1 x

1−

√1 - x2

√ (b) tan(cos

−1

x) =

1 − x2 x

√1 - x2

cos-1 x x

x) =

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136

Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Chapter 4

√

1 + x2 x

(d) sin(tan−1 x) = √

x 1 + x2

1 + x2

tan-1 x

13. (a) cos(tan−1 x) = √

√

1 1 + x2

(b) tan(cos−1 x) =

1 – x2

cos-1 x

tan-1 x

x) =

x2 − 1 x

(d) cot(sec−1 x) = √

√ −1

x2 - 1

sec-1 x

x x2 – 1 sec-1 x 1

−1.00 −0.80 −0.6 −0.40 −0.20 0.00 0.20 0.40 0.60 0.80 1.00 x sin−1 x −1.57 −0.93 −0.64 −0.41 −0.20 0.00 0.20 0.41 0.64 0.93 1.57 cos−1 x 3.14 2.50 2.21 1.98 1.77 1.57 1.37 1.16 0.93 0.64 0.00

(b)

(c)

y 3 2 1 x

0.5

uh a

1 x2 − 1

15. (a)

1 − x2 x

1 + x2

14. (a)

√1 + x2

us uf

-10

-1

c/2

5 x 10

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Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Exercise Set 4.4

137

16. (a) y = cot−1 x; if x > 0 then 0 < y < π/2 and x = cot y, tan y = 1/x, y = tan−1 (1/x);

us uf

if x < 0 then π/2 < y < π and x = cot y = cot(y − π), tan(y − π) = 1/x, y = π + tan−1 (b) y = sec−1 x, x = sec y, cos y = 1/x, y = cos−1 (1/x) (c) y = csc−1 x, x = csc y, sin y = 1/x, y = sin−1 (1/x) 17. (a) 55.0◦

(b) 33.6◦

1 x

(b) The domain of cot−1 x is (−∞, +∞), the range is (0, π); the domain of csc−1 x is (−∞, −1] ∪ [1, +∞), the range is [−π/2, 0) ∪ (0, π/2].

18. (a) Let x = f (y) = cot y, 0 < y < π, −∞ < x < +∞. √ Then f is diﬀerentiable and one-to-one x2 + 1 = − x2 + 1 = 0, and and f (f −1 (x)) = cot(cot−1 x) cos(cot−1 x) = −x x 1 d = lim −1 = − lim x2 + 1 = −1. [cot−1 x] x→0 f (f x→0 (x)) dx x=0

(b) If x = 0 then, from Exercise 16(a), d 1 1 1 1 d cot−1 x = tan−1 = − 2 = −√ . For x = 0, Part (a) shows the x dx dx x x2 + 1 1 + (1/x)2 1 d . same; thus for −∞ < x < +∞, [cot−1 x] = − √ 2 dx x +1 1 d du [cot−1 u] = − √ . 2 dx u + 1 dx

1 1 −1 d √ [csc−1 x] = − 2 = dx x |x| x2 − 1 1 − (1/x)2

(b) By the chain rule,

d du d −1 du √ [csc−1 u] = [csc−1 u] = 2 dx dx du |u| u − 1 dx

19. (a) By the chain rule,

20. (a) x = π − sin−1 (0.37) ≈ 2.7626 rad

(b) θ = 180◦ + sin−1 (0.61) ≈ 217.6◦

21. (a) x = π + cos−1 (0.85) ≈ 3.6964 rad

(b) θ = − cos−1 (0.23) ≈ −76.7◦

22. (a) x = tan−1 (3.16) − π ≈ −1.8773

(b) θ = 180◦ − tan−1 (0.45) ≈ 155.8◦

1−

23. (a)

x2 /9

(1/3) = 1/

9 − x2

(b) −2/ (b) −

√

|x|7

uh a

26. (a) y = 1/ tan x = cot x, dy/dx = − csc2 x −1

(b) y = (tan

27. (a)

1−

1/x2

−1

, dy/dx = −(tan

(−1/x2 ) = −

1 − (2x + 1)2

1 1+x

1 −1/2 x 2

=−

1 √ 2(1 + x) x

√ (b) −1/ e2x − 1

1 7 √ (7x6 ) = 14 x −1 |x| x14 − 1

25. (a)

24. (a) 2x/(1 + x4 )

−2

1 √ |x| x2 − 1

1 1 + x2

(b)

√

sin x sin x = = | sin x| 1 − cos2 x

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1, sin x > 0 −1, sin x < 0

Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Chapter 4

29. (a)

(cos−1

1 √ x) 1 − x2

1 (b) − √ −1 2 cot x(1 + x2 )

ex √ + ex sec−1 x |x| x2 − 1

(b)

30. (a) 0

3x2 (sin−1 x)2 √ + 2x(sin−1 x)3 1 − x2

(b) 0

31. x3 + x tan−1 y = ey , 3x2 +

28. (a) −

us uf

138

x (3x2 + tan−1 y)(1 + y 2 ) −1 y y + tan y = e y , y = 1 + y2 (1 + y 2 )ey − x

1 1 (xy + y) = − (1 − y ), 32. sin−1 (xy) = cos−1 (x − y), 2 2 1 − (x − y)2 1−x y y 1 − (x − y)2 + 1 − x2 y 2 y = 1 − x2 y 2 − x 1 − (x − y)2 (b)

c/2

33. (a)

c/2

x 0.5

-0.5 – c/2

– c/2

34. (a) sin−1 0.9 > 1, so it is not in the domain of sin−1 x (b) −1 ≤ sin−1 x ≤ 1 is necessary, or −0.841471 ≤ x ≤ 0.841471 R 6378 = sin−1 ≈ 23◦ R+h 16, 378 36. (a) If γ = 90◦ , then sin γ = 1, 1 − sin2 φ sin2 γ = 1 − sin2 φ = cos φ, D = tan φ tan λ = (tan 23.45◦ )(tan 65◦ ) ≈ 0.93023374 so h ≈ 21.1 hours. (b) If γ = 270◦ , then sin γ = −1, D = − tan φ tan λ ≈ −0.93023374 so h ≈ 2.9 hours.

35. (b) θ = sin−1

37. sin 2θ = gR/v 2 = (9.8)(18)/(14)2 = 0.9, 2θ = sin−1 (0.9) or 2θ = 180◦ − sin−1 (0.9) so θ = 12 sin−1 (0.9) ≈ 32◦ or θ = 90◦ − 12 sin−1 (0.9) ≈ 58◦ . The ball will have a lower parabolic trajectory for θ = 32◦ and hence will result in the shorter time of ﬂight.

38. 42 = 22 + 32 − 2(2)(3) cos θ, cos θ = −1/4, θ = cos−1 (−1/4) ≈ 104◦ √ 2 39. y = 0 when x2 = 6000v 60/g = 1000 30 for v = 400 and g = 32; √ /g, x = 10v √ tan θ = 3000/x = 3/ 30, θ = tan−1 (3/ 30) ≈ 29◦ . x x and cot β = so a+b b x −1 x −1 θ = cot − cot a+b b 1 1 1 1 dθ =− − (b) 2 2 dx a + b 1 + x /(a + b) b 1 + (x/b)2

uh a

40. (a) θ = α − β, cot α =

=−

b a+b − 2 2 2 (a + b) + x b + x2

␪ ␤

␣

which is negative for all x. Thus θ is a decreasing function of x, and it has no maximum since lim θ = +∞. x→0+

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Exercise Set 4.5

139

us uf

41. (a) Let θ = sin−1 (−x) then sin θ = −x, −π/2 ≤ θ ≤ π/2. But sin(−θ) = − sin θ and −π/2 ≤ −θ ≤ π/2 so sin(−θ) = −(−x) = x, −θ = sin−1 x, θ = − sin−1 x. (b) proof is similar to that in Part (a) 42. (a) Let θ = cos−1 (−x) then cos θ = −x, 0 ≤ θ ≤ π. But cos(π − θ) = − cos θ and 0 ≤ π − θ ≤ π so cos(π − θ) = x, π − θ = cos−1 x, θ = π − cos−1 x

(b) Let θ = sec−1 (−x) for x ≥ 1; then sec θ = −x and π/2 < θ ≤ π. So 0 ≤ π − θ < π/2 and π − θ = sec−1 sec(π − θ) = sec−1 (− sec θ) = sec−1 x, or sec−1 (−x) = π − sec−1 x. x (see ﬁgure) 1 − x2

43. (a) sin−1 x = tan−1 √

tan(tan−1 x + tan−1 y) =

x+y 1 − xy

(b) 2 tan−1 2 tan−1

1 1 1/2 + 1/3 + tan−1 = tan−1 = tan−1 1 = π/4 2 3 1 − (1/2) (1/3) 1 1 1 1/3 + 1/3 3 = tan−1 + tan−1 = tan−1 = tan−1 , 3 3 3 1 − (1/3) (1/3) 4

45. (a) tan−1

1 1 3 1 3/4 + 1/7 + tan−1 = tan−1 + tan−1 = tan−1 = tan−1 1 = π/4 3 7 4 7 1 − (3/4) (1/7)

x) = sin(cos

−1

(1/x)) =

√ 2 x2 − 1 1 1− = x |x|

46. sin(sec

−1

EXERCISE SET 4.5

2 x2 − 4 (x − 2)(x + 2) x+2 = lim = = lim x→2 x2 + 2x − 8 x→2 (x + 4)(x − 2) x→2 x + 4 3 lim

1. (a)

uh a

5 2 − lim 2x − 5 2 x→+∞ x = (b) lim = 7 x→+∞ 3x + 7 3 3 + lim x→+∞ x

2. (a)

(b)

√1 - x2

x+y tan(tan−1 x) + tan(tan−1 y) = 1 − xy 1 − tan(tan−1 x) tan(tan−1 y)

tan−1 x + tan−1 y = tan−1

sin-1 x

tan α + tan β , 1 − tan α tan β

44. tan(α + β) =

x 1 − x2

(b) sin−1 x + cos−1 x = π/2; cos−1 x = π/2 − sin−1 x = π/2 − tan−1 √

sin x sin x cos x = sin x = cos x so lim = lim cos x = 1 x→0 tan x x→0 tan x sin x (x − 1)(x + 1) x+1 2 x2 − 1 x2 − 1 = = 2 so lim 3 = 3 2 x→1 x − 1 x −1 (x − 1)(x + x + 1) x +x+1 3

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Chapter 4

lim

1/x =1 x→1 1

lim

ex =1 x→0 cos x

sec2 θ =1 θ→0 1

8. lim

13.

14.

1/x =0 x→+∞ 1 lim

12.

lim

x→0+

9e3x 3e3x = +∞ = lim x→+∞ 2 x→+∞ 2x

−x −1 − csc2 x = −∞ = lim+ = lim+ 2 1/x x→0 sin x x→0 2 sin x cos x

lim

−1/x x = lim+ 1/x = 0 x→0 e (−1/x2 )e1/x

x→0+

cos x = +∞ 2x

lim

x→0+

us uf

10.

11.

cos x = −1 1

x→π +

1 = 1/5 6x − 13

tet + et = −1 t→0 −et

lim

x→3

2 cos 2x = 2/5 x→0 5 cos 5x lim

140

(100)(99)x98 (100)(99)(98) · · · (1) 100x99 = lim = · · · = lim =0 x→+∞ x→+∞ x→+∞ ex ex ex √ cos x/ sin x 2/ 1 − 4x2 2 = lim cos x = 1 =2 17. lim 16. lim+ x→0 1 x→0 sec2 x/ tan x x→0+

21.

22.

23.

x→π

x→+∞

lim tan x ln x = lim + +

x→0

lim

x→0

x→(π/2)−

x→+∞

x 1 = lim x = 0 x→+∞ e ex

x−π 1 = lim = −2 x→π cot(x/2) −(1/2) csc2 (x/2)

ln x 1/x − sin2 x −2 sin x cos x = lim+ = lim = lim+ =0 cot x x→0 − csc2 x x→0+ x 1 x→0

sec 3x cos 5x =

lim (x − π) cot x = lim

x→π

lim xe−x = lim

x→+∞

sin(π/x) (−π/x2 ) cos(π/x) = lim π cos(π/x) = π = lim x→+∞ x→+∞ x→+∞ 1/x −1/x2

lim x sin(π/x) = lim

24.

lim (x − π) tan(x/2) = lim

x→π

19.

20.

lim

x→0

18.

1 1 1 1 + x2 = lim = x→0 3(1 + x2 ) 3x2 3

1−

lim

15.

x→π

lim

x→(π/2)−

5 cos 5x −5 sin 5x −5(+1) =− = lim = cos 3x x→(π/2)− −3 sin 3x (−3)(−1) 3

x−π 1 =1 = lim x→π sec2 x tan x

uh a

25. y = (1 − 3/x)x , lim ln y = lim x→+∞

x→+∞

26. y = (1 + 2x)−3/x , lim ln y = lim − x→0

x→0

ln(1 − 3/x) −3 = lim = −3, lim y = e−3 x→+∞ x→+∞ 1/x 1 − 3/x

3 ln(1 + 2x) 6 = lim − = −6, lim y = e−6 x→0 x→0 x 1 + 2x

ln(ex + x) ex + 1 = lim x = 2, lim y = e2 x→0 x→0 e + x x→0 x

27. y = (ex + x)1/x , lim ln y = lim x→0

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Exercise Set 4.5

141

b ln(1 + a/x) ab = lim = ab, lim y = eab x→+∞ x→+∞ 1 + a/x x→+∞ 1/x

x→+∞

x→1

ln cos(2/x) (−2/x2 )(− tan(2/x)) = lim x→+∞ x→+∞ 1/x2 −2/x3

30. y = [cos(2/x)]x , lim ln y = lim x→+∞

us uf

ln(2 − x) 2 sin2 (πx/2) = lim = 2/π, lim y = e2/π x→1 cot(πx/2) x→1 x→1 π(2 − x)

29. y = (2 − x)tan(πx/2) , lim ln y = lim

28. y = (1 + a/x)bx , lim ln y = lim

− tan(2/x) (2/x2 ) sec2 (2/x) = lim = −2, lim y = e−2 x→+∞ x→+∞ x→+∞ −1/x2 1/x 1 x − sin x 1 − cos x sin x 1 =0 = lim 31. lim − = lim = lim x→0 x sin x x→0 x cos x + sin x x→0 2 cos x − x sin x x→0 sin x x lim

x→0

9 1 − cos 3x 3 sin 3x 9 = lim cos 3x = = lim 2 x→0 x→0 2 2 x 2x

32.

= lim

33.

x 1 (x2 + x) − x2 √ = 1/2 = lim √ = lim 2 2 x→+∞ x→+∞ x +x+x x + x + x x→+∞ 1 + 1/x + 1

34.

ex − 1 ex ex − 1 − x = lim x = 1/2 = lim x x x x→0 xe + e − 1 x→0 xe + 2ex x→0 xe − x

lim

lim [x − ln(x2 + 1)] = lim [ln ex − ln(x2 + 1)] = lim ln

x→+∞

x→+∞ x

36.

e e e = lim = lim = +∞ so lim [x − ln(x2 + 1)] = +∞ x→+∞ x2 + 1 x→+∞ 2x x→+∞ 2

lim ln

x→+∞

x 1 = lim ln = ln(1) = 0 x→+∞ 1+x 1/x + 1

lim

x→+∞

ex , x2 + 1

35.

lim

ln x 1/x 1 = lim = lim =0 n n−1 x→+∞ x x→+∞ nx x→+∞ nxn xn nxn−1 (b) lim = lim = lim nxn = +∞ x→+∞ ln x x→+∞ 1/x x→+∞ lim

38. (a)

3x2 − 2x + 1 0 because it is not a form. x→1 3x2 − 2x 0

39. (a) L’Hˆ opital’s Rule does not apply to the problem lim 3x2 − 2x + 1 =2 x→1 3x2 − 2x

lim

x→+∞

1/(x ln x) 2 √ = lim √ =0 x→+∞ 1/(2 x) x ln x

uh a

41.

4x3 − 12x2 + 12x − 4 12x2 − 24x + 12 24x − 24 =0 = lim = lim 3 2 x→1 4x − 9x + 6x − 1 x→1 12x2 − 18x + 6 x→1 24x − 18 lim

40.

(b) lim

0.15

100

10000 0

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142

Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Chapter 4

42. y = xx , lim ln y = lim

x→0+

ln x = lim −x = 0, lim y = 1 1/x x→0+ x→0+

us uf

x→0+

0.5

43. y = (sin x)3/ ln x ,

3 ln sin x x = lim (3 cos x) lim+ ln y = lim+ = 3, + ln x sin x x→0 x→0 x→0 lim y = e3

x→0+

0.5

x→π/2−

4 sec2 x 4 = lim =4 sec x tan x x→π/2− sin x

45. ln x − ex = ln x −

1 e−x

lim e−x ln x = lim

x→+∞

e−x ln x − 1 ; e−x

0 0

e−x ln x − 1 = −∞ x→+∞ e−x

x→+∞

lim [ln ex − ln(1 + 2ex )] = lim ln x→+∞

x→+∞

-16

ex 1 + 2ex

-0.6 0

1 1 = ln ; +2 2

e−x

= lim ln x→+∞

ln x 1/x = lim = 0 by L’Hˆ opital’s Rule, x x→+∞ e ex

so lim [ln x − ex ] = lim

46.

lim

44.

horizontal asymptote y = − ln 2

uh a

47. y = (ln x)1/x ,

lim ln y = lim

x→+∞

-1.2

1.02

ln(ln x) 1 = lim = 0; x→+∞ x ln x x

lim y = 1, y = 1 is the horizontal asymptote

x→+∞

100

10000 1

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Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Exercise Set 4.5

48. y =

x+1 x+2

ln , lim ln y = lim x→+∞

x→+∞

x+1 x+2 1/x

−x2 = −1; x→+∞ (x + 1)(x + 2)

us uf

= lim

lim y = e−1 is the horizontal asymptote

x→+∞

(b) +∞

49. (a) 0

(d) −∞

50. (a) Type 00 ; y = x(ln a)/(1+ln x) ; lim+ ln y = lim+ x→0

x→0

143

(e) +∞

(f ) −∞

(ln a) ln x (ln a)/x = lim ln a = ln a, = lim 1/x 1 + ln x x→0+ x→0+

lim y = eln a = a

x→0+

(b) Type ∞0 ; same calculation as Part (a) with x → +∞ (ln a) ln(x + 1) ln a = lim = ln a, (c) Type 1∞ ; y = (x + 1)(ln a)/x , lim ln y = lim x→0 x→0 x→0 x + 1 x lim y = eln a = a x→0

lim

x→+∞

2 − cos x 2x − sin x 2 − (sin x)/x 2 does not exist, nor is it ±∞; lim = lim = x→+∞ x→+∞ 3 + cos x 3x + sin x 3 + (sin x)/x 3

52.

sin 2x 1 + 2 cos 2x x + sin 2x does not exist, nor is it ±∞; lim = lim =1 1+ 51. lim x→+∞ x→+∞ x→+∞ 1 x x

x(2 + sin 2x) 2 + sin 2x = lim , x→+∞ x+1 1 + 1/x which does not exist because sin 2x oscillates between −1 and 1 as x → +∞ sin x 1 1 + cos x + does not exist, nor is it ±∞; 54. lim x→+∞ x 2 2x lim (2 + x cos 2x + sin 2x) does not exist, nor is it ±∞; lim

lim

x→+∞

55.

lim+

R→0

x→+∞

x(2 + sin x) 2 + sin x = lim =0 2 x→+∞ x +1 x + 1/x V t −Rt/L L e

53.

Vt L

π/2 − x −1 = lim sin2 x = 1 = lim x→π/2 x→π/2 cot x x→π/2 − csc2 x x→π/2 sin x 1 1 cos x − (π/2 − x) sin x − tan x = lim − = lim (b) lim cos x (π/2 − x) cos x x→π/2 π/2 − x x→π/2 π/2 − x x→π/2 lim (π/2 − x) tan x = lim

uh a

56. (a)

= lim

−(π/2 − x) cos x −(π/2 − x) sin x − cos x

= lim

(π/2 − x) sin x + cos x =0 −(π/2 − x) cos x + 2 sin x

x→π/2

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Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Chapter 4

kt − 1 (ln k)k t = lim = ln k + x→+∞ t 1 t→0 t→0 √ (c) ln 0.3 = −1.20397, 1024 1024 0.3 − 1 = −1.20327; √ ln 2 = 0.69315, 1024 1024 2 − 1 = 0.69338

58. (a) No; sin(1/x) oscillates as x → 0.

lim x(k 1/x − 1) = lim+

57. (b)

(b)

0.05

0.35

-0.35

us uf

144

-0.05

(c) For the limit as x → 0+ use the Squeezing Theorem together with the inequalities −x2 ≤ x2 sin(1/x) ≤ x2 . For x → 0− do the same; thus lim f (x) = 0. 59. If k = −1 then lim (k + cos 3x) = k + 1 = 0, so lim x→0

k + cos 3x = ±∞. Hence k = −1, and by the x2

rule

x→0

√ −1 + cos 3x −3 sin 3x −32 cos 3x 32 = lim 2. = lim = − = −4 if 3 = ±2 x→0 x→0 x→0 2x 2 2 x2 lim

x→0

− cos(1/x) + 2x sin(1/x) which does not exist (nor is it ±∞). cos x

60. (a) Apply the rule to get lim

61.

lim

x→0+

x x 1 (b) Rewrite as lim [x sin(1/x)], but lim = lim = 1 and lim x sin(1/x) = 0, x→0 sin x x→0 sin x x→0 cos x x→0

x [x sin(1/x)] = (1)(0) = 0 thus lim x→0 sin x sin(1/x) sin x , lim = 1 but lim+ sin(1/x) does not exist because sin(1/x) oscillates between (sin x)/x x→0+ x x→0

−1 and 1 as x → +∞, so lim+

x→0

x sin(1/x) does not exist. sin x

f (g(x)) = x for all x in the domain of g, and g(f (x)) = x for all x in the domain of f . They are reflections of each other through the line y = x. The domain of one is the range of the other and vice versa. The equation y = f (x) can always be solved for x as a function of y. Functions with no inverses include y = x2 , y = sin x. (e) Yes, g is continuous; this is evident from the statement about the graphs in Part (b) above. (f ) Yes, g must be differentiable (where f = 0); this can be inferred from the graphs. Note that if f = 0 at a point then g cannot exist (infinite slope).

uh a

1. (a) (b) (c) (d)

CHAPTER 4 SUPPLEMENTARY EXERCISES

2. (a) For sin x, −π/2 ≤ x ≤ π/2; for cos x, 0 ≤ x ≤ π; for tan x, −π/2 < x < π/2; for sec x, 0 ≤ x < π/2 or π/2 < x ≤ π.

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Chapter 4 Supplementary Exercises

(b)

y = cos-1 x

y = sin-1 x

145

y = sin x

-1 -1

2 c/2

y = sec x

y = sec-1 x tan -1

y = sec-1 x x

x c/2

-1

-2

3. (a) x = f (y) = 8y 3 − 1; y = f −1 (x) =

x+1 8

1/3 =

y = sec x

1 (x + 1)1/3 2

- c/2

y = cos x

y = tan x

us uf

x c/2

(b) f (x) = (x − 1)2 ; f does not have an inverse because f is not one-to-one, for example f (0) = f (2) = 1. √ (c) x = f (y) = (ey )2 + 1; y = f −1 (x) = ln x − 1 = 12 ln(x − 1) x+2 y+2 ; y = f −1 (x) = y−1 x−1

(d) x = f (y) =

ad − bc ; if ad − bc = 0 then the function represents a horizontal line, no inverse. (cx + d)2 ay + b then If ad − bc = 0 then f (x) > 0 or f (x) < 0 so f is invertible. If x = f (y) = cy + d b − xd y = f −1 (x) = . xc − a

4. f (x) =

5. 3 ln e2x (ex )3 + 2 exp(ln 1) = 3 ln e2x + 3 ln(ex )3 + 2 · 1 = 3(2x) + (3 · 3)x + 2 = 15x + 2

6. Draw equilateral triangles of sides 5, 12, 13, and 3, 4, 5. Then sin[cos−1 (4/5)] = 3/5, sin[cos−1 (5/13)] = 12/13, cos[sin−1 (4/5)] = 3/5, cos[sin−1 (5/13)] = 12/13

(a) cos[cos−1 (4/5) + sin−1 (5/13)] = cos(cos−1 (4/5)) cos(sin−1 (5/13)) − sin(cos−1 (4/5)) sin(sin−1 (5/13)) 33 4 12 3 5 − = . 5 13 5 13 65

uh a

(b) sin[sin−1 (4/5) + cos−1 (5/13)] = sin(sin−1 (4/5)) cos(cos−1 (5/13)) + cos(sin−1 (4/5)) sin(cos−1 (5/13)) =

3 12 56 4 5 + = . 5 13 5 13 65

7. (a) f (x) = −3/(x + 1)2 . If x = f (y) = 3/(y + 1) then y = f −1 (x) = (3/x) − 1, so d −1 3 1 (f −1 (x) + 1)2 (3/x)2 3 =− f (x) = − 2 ; and −1 =− = − 2. dx x f (f (x)) 3 3 x

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146

Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Chapter 4

us uf

d −1 2 f (x) = ; (b) f (x) = ex/2 , f (x) = 12 ex/2 . If x = f (y) = ey/2 then y = f −1 (x) = 2 ln x, so dx x −1 1 2 = 2e−f (x)/2 = 2e− ln x = 2x−1 = and −1 x f (f (x)) 8. Y = ln(Cekt ) = ln C + ln ekt = ln C + kt, a line with slope k and Y -intercept ln C y

9. (a) 2

x 4

-2

(b)

c/2

(d)

x 1

y c/2 x 1

-c/2

(c)

c/2

10. (a)

(b) The curve y = e−x/2 sin 2x has x-intercepts at x = −π/2, 0, π/2, π, 3π/2. It intersects the curve y = e−x/2 at x = π/4, 5π/4, and it intersects the curve y = −e−x/2 at x = −π/4, 3π/4.

11. (a) The function ln x − x0.2 is negative at x = 1 and positive at x = 4, so it must be zero in between (IVT). (b) x = 3.654 ln x 1 = . The steps are reversible. x k (b) By zooming it is seen that the maximum value of y is approximately 0.368 (actually, 1/e), so there are two distinct solutions of xk = ex whenever k > 1/0.368 ≈ 2.717.

12. (a) If xk = ex then k ln x = x, or

x 2

-2

uh a

dy/dx dy = 1.07, or = 1.07y y dx

14. ln y = 2x ln 3 + 7x ln 5;

dy dy/dx = 2 ln 3 + 7 ln 5, or = (2 ln 3 + 7 ln 5)y y dx

13. ln y = ln 5000 + 1.07x;

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Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Chapter 4 Supplementary Exercises

us uf

(d) ln y =

abe−x (b) y = 15. (a) y = x3 + 1 so y = 3x2 . (1 + be−x )2 1 1 (c) y = ln x + ln(x + 1) − ln sin x + ln cos x, so 2 3 1 cos x 1 sin x 5x + 3 + − − cot x − tan x. y = − = 2x 3(x + 1) sin x cos x 6x(x + 1)

147

ln(1 + x) y 1 x/(1 + x) − ln(1 + x) ln(1 + x) = , , = − x y x2 x(1 + x) x2

y = ln

(1 + ex + e2x ) ex x dy = − ln(1 − e ), = (1 − ex )(1 + ex + e2x ) dx 1 − ex

(f )

dy 1 (1 + x)(1/x) = (1 + x)(1/x)−1 − ln(1 + x) x2 dx x

x x 1 1 y dy ex x x x + ln x = ex xe −1 + xe ln x (e) ln y = e ln x, =e + ln x , =x e x y x dx

16. y = aeax sin bx + beax cos bx and y = (a2 − b2 )eax sin bx + 2abeax cos bx, so y − 2ay + (a2 + b2 )y = (a2 − b2 )eax sin bx + 2abeax cos bx − 2a(aeax sin bx + beax cos bx) + (a2 + b2 )eax sin bx = 0. √ √ 17. sin(tan−1 x) = x/ 1 + x2 and cos(tan−1 x) = 1/ 1 + x2 , and y = 1 −2x x + 2√ = 0. (1 + x2 )2 1 + x2 (1 + x2 )3/2

y + 2 sin y cos3 y =

1 −2x , y = , hence 1 + x2 (1 + x2 )2

18. (a) Find x when y = 5 · 12 = 60 in. Since y = log x, x = 10y = 1060 in. This is approximately 2.68 × 1042 light-years, so even in astronomical terms it is a fabulously long distance.

(b) Find x when y = 100(5280)(12) in. Since y = 10x , x = log y = 6.80 in or 0.57 ft, approximately. 1 19. Set y = logb x and solve y = 1: y = = 1 x ln b 1 so x = . The curves intersect when (x, x) lies ln b on the graph of y = logb x, so x = logb x. From ln x from which Formula (9), Section 4.2, logb x = ln b 1/e ln x = 1, x = e, ln b = 1/e, b = e ≈ 1.4447.

2 x

uh a

√ 20. (a) Find the point of intersection: f (x) = x + k = ln x. The 1 1 √ slopes are equal, so m1 = = m2 = √ , x = 2, x = 4. x 2 x √ Then ln 4 = 4 + k, k = ln 4 − 2. √ k 1 (b) Since the slopes are equal m1 = √ = m2 = , so k x = 2. x √ 2 x At the point of intersection k x = ln x, 2 = ln x, x = e2 , k = 2/e.

y 2 x 2

y 2 x 0

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148

Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Chapter 4

dx dy dy dx dx dy =3 given y = x ln x. Then = = (1 + ln x) , so 1 + ln x = 3, ln x = 2, dt dt dt dx dt dt x = e2 .

us uf

21. Solve

22. Let P (x0 , y0 ) be a point on y = e3x then y0 = e3x0 . dy/dx = 3e3x so mtan = 3e3x0 at P and an equation of the tangent line at P is y − y0 = 3e3x0 (x − x0 ), y − e3x0 = 3e3x0 (x − x0 ). If the line passes through the origin then (0, 0) must satisfy the equation so −e3x0 = −3x0 e3x0 which gives x0 = 1/3 and thus y0 = e. The point is (1/3, e).

dβ dI

(c)

dβ dI

= I=10I0

1 db/W/m2 I0 ln 10

= I=100I0

1 100I0 ln 10

(b)

db/W/m2

100

dβ dI

1 db/W/m2 10I0 ln 10

I=100I0

8 20

25. (a)

(a)

dβ 10 = dI I ln 10

24. β = 10 log I − 10 log I0 ,

qk0 q(T − T0 ) q(T − T0 ) q dk − 2 = − 2 exp − = k0 exp − 2T 2T 2T0 T dT 2T0 T

23.

(b) as t tends to +∞, the population tends to 19 95 95 95 = = lim P (t) = lim = 19 −t/4 t→+∞ t→+∞ 5 − 4e−t/4 5 5 − 4 lim e t→+∞

26. (a)

-80

uh a

(b) P tends to 12 as t gets large; lim P (t) = lim t→+∞

t→+∞

60 60 60 = = 12 −t = −t 5 + 7e 5 5 + 7 lim e t→+∞

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Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Chapter 4 Supplementary Exercises

149

us uf

3.2

27. (b)

(c)

1 1 dy dy dy = − so < 0 at x = 1 and > 0 at x = e 2 x dx dx dx

4 2 x 2

(d) The slope is a continuous function which goes from a negative value to a positive value; therefore it must take the value zero in between, by the Intermediate Value Theorem. dy = 0 when x = 2 dx

(e)

28. In the case +∞ − (−∞) the limit is +∞; in the case −∞ − (+∞) the limit is −∞, because large positive (negative) quantities are added to large positive (negative) quantities. The cases +∞ − (+∞) and −∞ − (−∞) are indeterminate; large numbers of opposite sign are subtracted, and more information about the sizes is needed.

29. (a) when the limit takes the form 0/0 or ∞/∞ (b) Not necessarily; only if lim f (x) = 0. Consider g(x) = x; lim g(x) = 0. For f (x) choose x→a

x→0

cos x x2 |x|1/2 does not exist, lim = 0, and lim = +∞. cos x, x2 , and |x|1/2 . Then: lim x→0 x x→0 x x→0 x2 ex ex ex = lim = +∞ = lim x→+∞ x→+∞ x→+∞ x2 x→+∞ 2x x→+∞ 2 so lim (ex /x2 − 1) = +∞ and thus lim x2 (ex /x2 − 1) = +∞ lim (ex − x2 ) = lim x2 (ex /x2 − 1), but lim x→+∞

30. (a)

ln x = x4 − 1

lim

x→1

1 ln x = x4 − 1 2

lim ax ln a = ln a

(c)

ln x 1/x 1 = lim (b) lim 4 = ; lim x→1 x − 1 x→1 4x3 4 x→1

x→+∞

uh a

x→0

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Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH CHAPTER 5

The Derivative in Graphing and Applications

1. (a) f > 0 and f > 0

(b) f > 0 and f < 0 y

(d) f < 0 and f < 0

(b)

(c)

2. (a)

us uf

EXERCISE SET 5.1

(d)

A: dy/dx < 0, d2 y/dx2 > 0 B: dy/dx > 0, d2 y/dx2 < 0 C: dy/dx < 0, d2 y/dx2 < 0

A: dy/dx < 0, d2 y/dx2 < 0 B: dy/dx < 0, d2 y/dx2 > 0 C: dy/dx > 0, d2 y/dx2 < 0

5. An inﬂection point occurs when f changes sign: at x = −1, 0, 1 and 2.

uh a

6. (a) f (0) < f (1) since f > 0 on (0, 1). (c) f (0) > 0 by inspection. (e) f (0) < 0 since f is decreasing there.

7. (a) [4, 6] (d) (2, 3) and (5, 7)

(b) f (1) > f (2) since f < 0 on (1, 2). (d) f (1) = 0 by inspection. (f ) f (2) = 0 since f has a minimum there.

(b) [1, 4] and [6, 7] (e) x = 2, 3, 5

150

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Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Exercise Set 5.1

(1, 2) (2, 3) (3, 4) (4, 5) (5, 6) (6, 7) − − − + + − + − + + − −

us uf

f f

9. (a) f is increasing on [1, 3] (c) f is concave up on (−∞, 2), (4, +∞) (e) points of inﬂection at x = 2, 4

(b) f is decreasing on (−∞, 1], [3, +∞] (d) f is concave down on (2, 4)

10. (a) f is increasing on (−∞, +∞) (c) f is concave up on (−∞, 1), (3, +∞) (e) f has points of inﬂection at x = 1, 3

(b) f is nowhere decreasing (d) f is concave down on (1, 3)

(a) [5/2, +∞) (c) (−∞, +∞) (e) none

(b) (−∞, 5/2] (d) none

12. f (x) = −2(x + 3/2) f (x) = −2

(a) (−∞, −3/2] (c) none (e) none

(b) [−3/2, +∞) (d) (−∞, +∞)

13. f (x) = 3(x + 2)2 f (x) = 6(x + 2)

(a) (−∞, +∞) (c) (−2, +∞) (e) −2

14. f (x) = 3(4 − x2 ) f (x) = −6x

(a) [−2, 2] (c) (−∞, 0) (e) 0

15. f (x) = 12x2 (x − 1) f (x) = 36x(x − 2/3)

(a) [1, +∞) (c) (−∞, 0), (2/3, +∞) (e) 0, 2/3

(b) (−∞, 1] (d) (0, 2/3)

(a) [−2, 0], [2,√ +∞) √ (c) (−∞, −2/ √ √3), (2/ 3, +∞) (e) −2/ 3, 2/ 3

(b) (−∞,√−2], [0, √ 2] (d) (−2/ 3, 2/ 3)

(x2

4x + 2)2

(a) [0, +∞)

17. f (x) =

11. f (x) = 2x − 5 f (x) = 2

16. f (x) = 4x(x2 − 4) f (x) = 12(x2 − 4/3)

f (x) = −4

(b) (−∞, 0]

2 − x2 2x(x2 − 6) f (x) = 2 2 (x + 2) (x2 + 2)3 √ √ √ √ (b) (−∞, − 2], [ 2, +∞) (a) [− 2, 2] √ √ √ √ (d) (−∞, − 6), (0, 6) (e) − 6, 0, 6

uh a

19. f (x) = 13 (x + 2)−2/3

f (x) =

− 29 (x

−5/3

+ 2)

(b) none (d) (−∞, −2)

(b) (−∞, −2], [2, +∞) (d) (0, +∞)

3x2 − 2 (x2 + 2)3

(d) (−∞, − 2/3), ( 2/3, +∞)

18. f (x) =

151

√ √ (c) (− 6, 0), ( 6, +∞)

(a) (−∞, +∞)

(b) none

(d) (−2, +∞)

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Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Chapter 5

4(x + 1) 3x2/3 4(x − 2) f (x) = 9x5/3

21. f (x) =

4(x − 1/4) 3x2/3 4(x + 1/2) f (x) = 9x5/3 2

(d) (−∞, 0), (0, +∞)

(a) [−1, +∞)

(b) (−∞, −1]

(d) (0, 2)

(b) (−∞, 0]

(e) 0, 2

22. f (x) =

23. f (x) = −xe−x

(a) [0, +∞)

(a) [1/4, +∞)

(b) (−∞, 1/4]

(d) (−1/2, 0)

f (x) =

− 29 x−4/3

(e) −1/2, 0

/2 2

f (x) = (−1 + x2 )e−x

(a) (−∞, 0] (c) (−∞, −1), (1, +∞) (e) −1, 1

24. f (x) = (2x2 + 1)ex

(a) (−∞, +∞) (c) (0, +∞) (e) 0

f (x) = 2x(2x2 + 3)ex 2x 1 + x2 1 − x2 f (x) = 2 (1 + x2 )2

(d) (−∞, −1), (1, +∞)

(a) [e−1/2 , +∞)

(b) (0, e−1/2 ]

(d) (0, e−3/2 )

(b) [0, π] (d) (0, π/2), (3π/2, 2π)

(a) [π, 2π] (c) (π/2, 3π/2) (e) π/2, 3π/2

27. f (x) = − sin x f (x) = − cos x

(b) (−∞, 0]

(e) −1, 1

26. f (x) = x(2 ln x + 1) f (x) = 2 ln x + 3

(b) none (d) (−∞, 0)

(a) [0, +∞)

25. f (x) =

(b) [0, +∞) (d) (−1, 1)

p 0

uh a

(0, π/4], [π/2, 3π/4] [π/4, π/2], [3π/4, π] (0, π/8), (3π/8, 5π/8), (7π/8, π) (π/8, 3π/8), (5π/8, 7π/8) π/8, 3π/8, 5π/8, 7π/8

-1

28. f (x) = 2 sin 4x f (x) = 8 cos 4x (a) (b) (c) (d) (e)

us uf

20. f (x) = 23 x−1/3

152

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Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Exercise Set 5.1

29. f (x) = sec2 x f (x) = 2 sec2 x tan x

153

(b) none (d) (−π/2, 0)

us uf

(a) (−π/2, π/2) (c) (0, π/2) (e) 0

-10

30. f (x) = 2 − csc2 x 2

(b) (0, π/4], [3π/4, π) (d) (π/2, π) 0

cos x f (x) = 2 csc x cot x = 2 3 sin x (a) [π/4, 3π/4] (c) (0, π/2) (e) π/2

-2

31. f (x) = cos 2x f (x) = −2 sin 2x (b) [π/4, 3π/4] (d) (0, π/2)

(a) [0, π/4], [3π/4, π] (c) (π/2, π) (e) π/2

0.5

-0.5

32. f (x) = −2 cos x sin x − 2 cos x = −2 cos x(1 + sin x) f (x) = 2 sin x (sin x + 1) − 2 cos2 x = 2 sin x(sin x + 1) − 2 + 2 sin2 x = 4(1 + sin x)(sin x − 1/2) Note: 1 + sin x ≥ 0

(c)

4 4

uh a y

(b)

(c)

(b)

34. (a)

-2

33. (a)

(b) [0, π/2], [3π/2, 2π] (d) (0, π/6), (5π/6, 2π)

(a) [π/2, 3π/2] (c) (π/6, 5π/6) (e) π/6, 5π/6

x 2

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154

Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Chapter 5

35. (a) f (x) = 3(x − a)2 , f (x) = 6(x − a); inﬂection point is (a, 0) (b) f (x) = 4(x − a)3 , f (x) = 12(x − a)2 ; no inﬂection points

2.5

37. f (x) = 1/3 − 1/[3(1 + x)2/3 ] so f is increasing √ on [0, +∞) 3 1 + x > 0, thus if x > 0, then f (x) > f (0) = 0, 1 + x/3 − √ 3 1 + x < 1 + x/3.

0 0

38. f (x) = sec2 x − 1 so f is increasing on [0, π/2) thus if 0 < x < π/2, then f (x) > f (0) = 0, tan x − x > 0, x < tan x.

us uf

36. For n ≥ 2, f (x) = n(n − 1)(x − a)n−2 ; there is a sign change of f (point of inﬂection) at (a, 0) if and only if n is odd. For n = 1, y = x − a, so there is no point of inﬂection.

39. x ≥ sin x on [0, +∞): let f (x) = x − sin x. Then f (0) = 0 and f (x) = 1 − cos x ≥ 0, so f (x) is increasing on [0, +∞).

-1

40. Let f (x) = 1 − x2 /2 − cos x for x ≥ 0. Then f (0) = 0 and f (x) = −x + sin x. By Exercise 35, f (x) ≤ 0 for x ≥ 0, so f (x) ≤ 0 for all x ≥ 0, that is, cos x ≥ 1 − x2 /2.

1.2

uh a

(c)

41. (a) Let f (x) = x − ln(x + 1) for x ≥ 0. Then f (0) = 0 and f (x) = 1 − 1/(x + 1) ≥ 0 for x ≥ 0, so f is increasing for x ≥ 0 and thus ln(x + 1) ≤ x for x ≥ 0. (b) Let g(x) = x − 12 x2 − ln(x + 1). Then g(0) = 0 and g (x) = 1 − x − 1/(x + 1) ≤ 0 for x ≥ 0 since 1 − x2 ≤ 1. Thus g is decreasing and thus ln(x + 1) ≥ x − 21 x2 for x ≥ 0.

2 0

42. (a) Let h(x) = ex − 1 − x for x ≥ 0. Then h(0) = 0 and h (x) = ex − 1 ≥ 0 for x ≥ 0, so h(x) is increasing. (b) Let h(x) = ex −1−x− 12 x2 . Then h(0) = 0 and h (x) = ex −1−x. By Part (a), ex −1−x ≥ 0 for x ≥ 0, so h(x) is increasing.

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Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Exercise Set 5.1

(c)

2 0

43. Points of inﬂection at x = −2, +2. Concave up on (−5, −2) and (2, 5); concave down on (−2, 2). Increasing on [−3.5829, 0.2513] and [3.3316, 5], and decreasing on [−5, −3.5829] and [0.2513, 3.3316].

-250 1

-5

√ 44. Points of inﬂection at x√ = ±1/ 3. Concave up √ on [−5, −1/ 3] √ and [1/ 3, 5], and concave down √ on [−1/ 3, 1/ 3]. Increasing on [−5, 0] and decreasing on [0, 5].

250

us uf

155

-5

-2

90x3 − 81x2 − 585x + 397 . The denominator has complex roots, so is always positive; (3x2 − 5x + 8)3 hence the x-coordinates of the points of inﬂection of f (x) are the roots of the numerator (if it changes sign). A plot of the numerator over [−5, 5] shows roots lying in [−3, −2], [0, 1], and [2, 3]. To six decimal places the roots are x = −2.464202, 0.662597, 2.701605.

45. f (x) = 2

2x5 + 5x3 + 14x2 + 30x − 7 . Points of inﬂection will occur when the numerator changes (x2 + 1)5/2 sign, since the denominator is always positive. A plot of y = 2x5 + 5x3 + 14x2 + 30x − 7 shows that there is only one root and it lies in [0, 1]. To six decimal place the point of inﬂection is located at x = 0.210970.

46. f (x) =

47. f (x1 ) − f (x2 ) = x21 − x22 = (x1 + x2 )(x1 − x2 ) < 0 if x1 < x2 for x1 , x2 in [0, +∞), so f (x1 ) < f (x2 ) and f is thus increasing. 1 1 x2 − x1 − = > 0 if x1 < x2 for x1 , x2 in (0, +∞), so f (x1 ) > f (x2 ) and x1 x2 x1 x2 thus f is decreasing.

48. f (x1 ) − f (x2 ) =

uh a

49. (a) If x1 < x2 where x1 and x2 are in I, then f (x1 ) < f (x2 ) and g(x1 ) < g(x2 ), so f (x1 ) + g(x1 ) < f (x2 ) + g(x2 ), (f + g)(x1 ) < (f + g)(x2 ). Thus f + g is increasing on I. (b) Case I: If f and g are ≥ 0 on I, and if x1 < x2 where x1 and x2 are in I, then 0 < f (x1 ) < f (x2 ) and 0 < g(x1 ) < g(x2 ), so f (x1 )g(x1 ) < f (x2 )g(x2 ), (f · g)(x1 ) < (f · g)(x2 ). Thus f · g is increasing on I. Case II: If f and g are not necessarily positive on I then no conclusion can be drawn: for example, f (x) = g(x) = x are both increasing on (−∞, 0), but (f · g)(x) = x2 is decreasing there.

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Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

156

Chapter 5

50. (a) f (x) = x, g(x) = 2x

(b) f (x) = x, g(x) = x + 6

us uf

b b , f (x) = 0 when x = − . f changes its direction of 51. (a) f (x) = 6ax + 2b = 6a x + 3a 3a b b concavity at x = − so − is an inﬂection point. 3a 3a

(b) If f (x) = ax3 + bx2 + cx + d has three x-intercepts, then it has three roots, say x1 , x2 and x3 , so we can write f (x) = a(x − x1 )(x − x2 )(x − x3 ) = ax3 + bx2 + cx + d, from which it b 1 follows that b = −a(x1 + x2 + x3 ). Thus − = (x1 + x2 + x3 ), which is the average. 3a 3 (c) f (x) = x(x2 − 3x2 + 2) = x(x − 1)(x − 2) so the intercepts are 0, 1, and 2 and the average is 1. f (x) = 6x − 6 = 6(x − 1) changes sign at x = 1.

b 52. f (x) = 6x + 2b, so the point of inﬂection is at x = − . Thus an increase in b moves the point of 3 inﬂection to the left.

53. (a) Let x1 < x2 belong to (a, b). If both belong to (a, c] or both belong to [c, b) then we have f (x1 ) < f (x2 ) by hypothesis. So assume x1 < c < x2 . We know by hypothesis that f (x1 ) < f (c), and f (c) < f (x2 ). We conclude that f (x1 ) < f (x2 ). (b) Use the same argument as in Part (a), but with inequalities reversed.

54. By Theorem 5.1.2, f is increasing on any interval [(2n−1)π, 2(n+1)π] (n = 0, ±1, ±2, · · ·), because f (x) = 1 + cos x > 0 on ((2n − 1)π, (2n + 1)π). By Exercise 53 (a) we can piece these intervals together to show that f (x) is increasing on (−∞, +∞).

55. By Theorem 5.1.2, f is decreasing on any interval [(2nπ + π/2, 2(n + 1)π + π/2] (n = 0, ±1, ±2, · · ·), because f (x) = − sin x + 1 < 0 on ((2nπ + π/2, 2(n + 1)π + π/2). By Exercise 53 (b) we can piece these intervals together to show that f (x) is decreasing on (−∞, +∞).

56. By zooming on the graph of y (t), maximum increase is at x = −0.577 and maximum decrease is at x = 0.577. y

57. 2

infl pt

58.

4 3

infl pts

2 1 x

LAke−kt LAk S, so y (0) = (1 + Ae−kt )2 (1 + A)2

59. (a) y (t) =

uh a

(b) The rate of growth increases to its maximum, which occurs when y is halfway between 0 and 1 L, or when t = ln A; it then decreases back towards zero. k dy is maximized when y lies half way between 0 and L, i.e. y = L/2. (c) From (2) one sees that dt This follows since the right side of (2) is a parabola (with y as independent variable) with 1 y-intercepts y = 0, L. The value y = L/2 corresponds to t = ln A, from (4). k

60. Find t so that N (t) is maximum. The size of the population is increasing most rapidly when t = 8.4 years.

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Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Exercise Set 5.2

61. t = 7.67

157

us uf

1000

15 0

62. Since 0 < y < L the right-hand side of (4) of Example 7 can change sign only if the factor L − 2y L L changes sign, which it does when y = L/2, at which point we have , 1 = Ae−kt , = 2 1 + Ae−kt 1 t = ln A. k

x→+∞

63. (a) g(x) has no zeros: There can be no zero of g(x) on the interval −∞ < x < 0 because if there were, say g(x0 ) = 0 where x0 < 0, then g (x) would have to be positive between x = x0 and x = 0, say g (x1 ) > 0 where x0 < x1 < 0. But then g (x) cannot be concave up on the interval (x1 , 0), a contradiction. There can be no zero of g(x) on 0 < x < 4 because g(x) is concave up for 0 < x < 4 and thus 2 the graph of g(x), for 0 < x < 4, must lie above the line y = − x + 2, which is the tangent 3 line to the curve at (0, 2), and above the line y = 3(x − 4) + 3 = 3x − 9 also for 0 < x < 4 (see ﬁgure). The ﬁrst condition says that g(x) could only be zero for x > 3 and the second condition says that g(x) could only be zero for x < 3, thus g(x) has no zeros for 0 < x < 4. Finally, if 4 < x < +∞, g(x) could only have a zero if g (x) were negative somewhere for x > 4, and since g (x) is decreasing there we would ultimately have g(x) < −10, a contradiction. (b) one, between 0 and 4 y (c) We must have lim g (x) = 0; if the limit were −5 then g(x) would at some time cross the line x = −10; if the limit were 5 then, since g is concave down for x > 4 and g (4) = 3, g must decrease for x > 4 and thus the limit would be < 4.

4 slope -2/3 1

EXERCISE SET 5.2 y

1. (a)

uh a

f (x)

(b)

f (x) x

(d)

f (x) x

f (x)

(c)

slope 4

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158

Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Chapter 5

2. (a)

(b)

(c)

us uf

(d)

3. (a) f (x) = 6x − 6 and f (x) = 6, with f (1) = 0. For the ﬁrst derivative test, f < 0 for x < 1 and f > 0 for x > 1. For the second derivative test, f (1) > 0. (b) f (x) = 3x2 − 3 and f (x) = 6x. f (x) = 0 at x = ±1. First derivative test: f > 0 for x < −1 and x > 1, and f < 0 for −1 < x < 1, so there is a relative maximum at x = −1, and a relative minimum at x = 1. Second derivative test: f < 0 at x = −1, a relative maximum; and f > 0 at x = 1, a relative minimum.

4. (a) f (x) = 2 sin x cos x = sin 2x (so f (0) = 0) and f (x) = 2 cos 2x. First derivative test: if x is near 0 then f < 0 for x < 0 and f > 0 for x > 0, so a relative minimum at x = 0. Second derivative test: f (0) = 2 > 0, so relative minimum at x = 0. (b) g (x) = 2 tan x sec2 x (so g (0) = 0) and g (x) = 2 sec2 x(sec2 x + 2 tan2 x). First derivative test: g < 0 for x < 0 and g > 0 for x > 0, so a relative minimum at x = 0. Second derivative test: g (0) = 2 > 0, relative minimum at x = 0. (c) Both functions are squares, and so are positive for values of x near zero; both functions are zero at x = 0, so that must be a relative minimum.

5. (a) f (x) = 4(x − 1)3 , g (x) = 3x2 − 6x + 3 so f (1) = g (1) = 0. (b) f (x) = 12(x − 1)2 , g (x) = 6x − 6, so f (1) = g (1) = 0, which yields no information. (c) f < 0 for x < 1 and f > 0 for x > 1, so there is a relative minimum at x = 1; g (x) = 3(x − 1)2 > 0 on both sides of x = 1, so there is no relative extremum at x = 1.

6. (a) f (x) = −5x4 , g (x) = 12x3 − 24x2 so f (0) = g (0) = 0. (b) f (x) = −20x3 , g (x) = 36x2 − 48x, so f (0) = g (0) = 0, which yields no information. (c) f < 0 on both sides of x = 0, so there is no relative extremum there; g (x) = 12x2 (x − 2) < 0 on both sides of x = 0 (for x near 0), so again there is no relative extremum there.

7. (a) f (x) = 3x2 + 6x − 9 = 3(x + 3)(x − 1), f (x) = 0 when x = −3, 1 (stationary points). √ (b) f (x) = 4x(x2 − 3), f (x) = 0 when x = 0, ± 3 (stationary points).

uh a

8. (a) f (x) = 6(x2 − 1), f (x) = 0 when x = ±1 (stationary points). (b) f (x) = 12x3 − 12x2 = 12x2 (x − 1), f (x) = 0 when x = 0, 1 (stationary points). √ 9. (a) f (x) = (2 − x2 )/(x2 + 2)2 , f (x) = 0 when x = ± 2 (stationary points). (b) f (x) = 23 x−1/3 = 2/(3x1/3 ), f (x) does not exist when x = 0. 10. (a) f (x) = 8x/(x2 + 1)2 , f (x) = 0 when x = 0 (stationary point). (b) f (x) = 13 (x + 2)−2/3 , f (x) does not exist when x = −2.

11. (a) f (x) = x = 0.

4(x + 1) , f (x) = 0 when x = −1 (stationary point), f (x) does not exist when 3x2/3

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Exercise Set 5.2

159

(b) f (x) = −3 sin 3x, f (x) = 0 when sin 3x = 0, 3x = nπ, n = 0, ±1, ±2, · · · x = nπ/3, n = 0, ±1, ±2, · · · (stationary points)

4(x − 3/2) , f (x) = 0 when x = 3/2 (stationary point), f (x) does not exist when 3x2/3

(b) f (x) = | sin x| =

sin x, sin x ≥ 0 − sin x, sin x < 0

cos x, sin x > 0 − cos x, sin x < 0

so f (x) =

us uf

12. (a) f (x) = x = 0.

and f (x) does

not exist when x = nπ, n = 0, ±1, ±2, · · · (the points where sin x = 0) because lim f (x) = lim f (x) (see Theorem preceding Exercise 75, Section 3.3). Now f (x) = 0 x→nπ +

x→nπ −

when ± cos x = 0 provided sin x = 0 so x = π/2+nπ, n = 0, ±1, ±2, · · · are stationary points.

13. (a) none (b) x = 1 because f changes sign from + to − there (c) none because f = 0 (never changes sign)

14. (a) x = 1 because f (x) changes sign from − to + there (b) x = 3 because f (x) changes sign from + to − there (c) x = 2 because f (x) changes sign there

16. (a) x = 1

15. (a) x = 2 because f (x) changes sign from − to + there. (b) x = 0 because f (x) changes sign from + to − there. (c) x = 1, 3 because f (x) changes sign at these points.

(b) x = 5

√ 17. (a) critical numbers x = 0, ± 5; f :

- - 0 + + 0 - - 0 + + - 5

√ x = 0: relative maximum; x = ± 5: relative minimum (b) critical number x = 1, −1; f :

+ + + 0 - - - 0 + + + -1

x = −1: relative maximum; x = 1: relative minimum

18. (a) critical numbers x = 0, −1/2, 1; f :

+ + + 0 - 0 - - - 0 + -1 2

x = 0: neither; x = −1/2: relative maximum; x = 1: relative minimum + + 0 - - ? + + 0 - -

(b) critical numbers: x = ±3/2, −1; f :

3 2

-1

3 2

uh a

x = ±3/2: relative maximum; x = −1: relative minimum

19. (a) critical point x = 0; f : x = 0: relative minimum (b) critical point x = ln 2: f : x = ln 2: relative minimum

-- - 0 + + + 0 -- - 0 + + + ln 2

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Chapter 5

- - -0+ + + 0- - -1

20. (a) critical points x = −1, 1:f : x = −1: relative minimum; x = 1: relative maximum (b) x = 1: neither

-- - 0 -- 1

21. f (x) = −2(x + 2); critical number x = −2; f (x):

+ + + 0 – – – -2

f (x) = −2; f (−2) < 0, f (−2) = 5; relative max of 5 at x = −2 22. f (x) = 6(x − 2)(x − 1); critical numbers x = 1, 2; f (x):

+ + + 0 - - - 0 + + + 1

us uf

160

23. f (x) = 2 sin x cos x = sin 2x; critical numbers x = π/2, π, 3π/2; f (x):

f (x) = 12x − 18; f (1) < 0, f (2) > 0, f (1) = 5, f (2) = 4; relative min of 4 at x = 2, relative max of 5 at x = 1 + + 0 - - 0 + + 0 - c

3c 2

c 2

f (x) = 2 cos 2x; f (π/2) < 0, f (π) > 0, f (3π/2) < 0, f (π/2) = f (3π/2) = 1, f (π) = 0; relative min of 0 at x = π, relative max of 1 at x = π/2, 3π/2 - - 0 + + + + + 0 - -

24. f (x) = 1/2 − cos x; critical numbers x = π/3, 5π/3; f (x):

c 3

5c 3

f (x) = − sin x; √ f (π/3) < 0, f (5π/3) >√0 f (π/3) = π/6 − 3/2, f (5π/3) √ √ = 5π/6 + 3/2; relative minimum of π/6 − 3/2 at x = π/3, relative maximum of 5π/6 + 3/2 at x = 5π/3

25. f (x) = 3x2 + 5; no relative extrema because there are no critical numbers. 26. f (x) = 4x(x2 − 1); critical numbers x = 0, 1, −1 f (x) = 12x2 − 4; f (0) < 0, f (1) > 0, f (−1) > 0 relative minimum of 6 at x = 1, −1, relative maximum of 7 at x = 0

27. f (x) = (x − 1)(3x − 1); critical numbers x = 1, 1/3 f (x) = 6x − 4; f (1) > 0, f (1/3) < 0 relative minimum of 0 at x = 1, relative maximum of 4/27 at x = 1/3

28. f (x) = 2x2 (2x + 3); critical numbers x = 0, −3/2 relative minimum of −27/16 at x = −3/2 (ﬁrst derivative test)

29. f (x) = 4x(1 − x2 ); critical numbers x = 0, 1, −1 f (x) = 4 − 12x2 ; f (0) > 0, f (1) < 0, f (−1) < 0 relative minimum of 0 at x = 0, relative maximum of 1 at x = 1, −1

uh a

30. f (x) = 10(2x − 1)4 ; critical number x = 1/2; no relative extrema (ﬁrst derivative test)

31. f (x) = 45 x−1/5 ; critical number x = 0; relative minimum of 0 at x = 0 (ﬁrst derivative test)

32. f (x) = 2 + 23 x−1/3 ; critical numbers x = 0, −1/27 relative minimum of 0 at x = 0, relative maximum of 1/27 at x = −1/27 33. f (x) = 2x/(x2 + 1)2 ; critical number x = 0; relative minimum of 0 at x = 0

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Exercise Set 5.2

161

34. f (x) = 2/(x + 2)2 ; no critical numbers (x = −2 is not in the domain of f ) no relative extrema

35. f (x) = 2x/(1 + x2 ); critical point at x = 0; relative minimum of 0 at x = 0 (ﬁrst derivative test)

us uf

36. f (x) = x(2 + x)ex ; critical points at x = 0, −2; relative minimum of 0 at x = 0 and relative maximum of 4/e2 at x = −2 (ﬁrst derivative test)

38. f (x) = −1 if x < 3, f (x) = 2x if x > 3, f (3) does not exist; critical number x = 3, relative minimum of 6 at x = 3

39. f (x) = 2 cos 2x if sin 2x > 0, f (x) = −2 cos 2x if sin 2x < 0, f (x) does not exist when x = π/2, π, 3π/2; critical numbers x = π/4, 3π/4, 5π/4, 7π/4, π/2, π, 3π/2 relative minimum of 0 at x = π/2, π, 3π/2; relative maximum of 1 at x = π/4, 3π/4, 5π/4, 7π/4

37. f (x) = 2x if |x| > 2, f (x) = −2x if |x| < 2, f (x) does not exist when x = ±2; critical numbers x = 0, 2, −2 relative minimum of 0 at x = 2, −2, relative maximum of 4 at x = 0

√ 40. f (x) = 3 + 2 cos x; critical numbers x = 5π/6, √ 7π/6 relative minimum of 7 √3π/6 − 1 at x = 7π/6; relative maximum of 5 3π/6 + 1 at x = 5π/6

41. f (x) = − sin 2x; critical numbers x = π/2, π, 3π/2 relative minimum of 0 at x = π/2, 3π/2; relative maximum of 1 at x = π

ad m m

42. f (x) = (2 cos x − 1)/(2 − cos x)2 ; critical numbers x = π/3, √ 5π/3 relative maximum of √ 3/3 at x = π/3, relative minimum of − 3/3 at x = 5π/3 0.8

uh a

-0.8

43. f (x) = ln x + 1, f (x) = 1/x; f (1/e) = 0, f (1/e) > 0; relative minimum of −1/e at x = 1/e

2.5

0 -0.5

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2.5

162

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Chapter 5

44. f (x) = −2

ex − e−x = 0 when x = 0. (ex + e−x )2

us uf

By the ﬁrst derivative test f (x) > 0 for x < 0 and f (x) < 0 for x > 0; relative maximum of 1 at x = 0

-2

45. f (x) = 2x(1 − x)e−2x = 0 at x = 0, 1. f (x) = (4x2 − 8x + 2)e−2x ; f (0) > 0 and f (1) < 0, so a relative minimum of 0 at x = 0 and a relative maximum of 1/e2 at x = 1.

0.14

-0.3

46. f (x) = 10/x − 1 = 0 at x = 10; f (x) = −10/x2 < 0; relative maximum of 10(ln(10) − 1) ≈ 13.03 at x = 10

47. Relative minima at x = −3.58, 3.33; relative maximum at x = 0.25 250

-5

-4

48. Relative minimum at x = −0.84; relative maximum at x = 0.84 1.2

-6

-250

49. Relative minimum at x = −1.20 and a relative maximum at x = 1.80

y f''(x)

-1.2

50. Relative maximum at x = −0.78 and a relative minimum at x = 1.55 y f''(x)

uh a

-4

-2

-1

f'(x) x 2

x -4 -2 f'(x)

-5

-2

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Exercise Set 5.2

x4 + 3x2 − 2x (x2 + 1)2 x3 − 3x2 − 3x + 1 (x2 + 1)3

f'(x)

f (x) = −2

y 2

us uf

51. f (x) =

163

Relative maximum at x = 0, relative minimum at x ≈ 0.59

–2 f''(x)

–2

4x3 − sin 2x 52. f (x) = √ , 2 x4 + cos2 x

f '(x)

6x2 − cos 2x (4x3 − sin 2x)(4x3 − sin 2x) f (x) = √ − 4(x4 + cos2 x)3/2 x4 + cos2 x

f''(x) –2

Relative minima at x ≈ ±0.62, relative maximum at x = 0

–2

k k 2x3 − k , then f (x) = 2x − 2 = . f has a relative extremum when x x x2 3 3 3 2x − k = 0, so k = 2x = 2(3) = 54. x k − x2 (b) Let f (x) = 2 . f has a relative extremum when k − x2 = 0, so , then f (x) = 2 x +k (x + k)2 k = x2 = 32 = 9.

54. (a) relative minima at x ≈ ±0.6436, relative maximum at x = 0 y

-0.5

(b) x ≈ ±0.6436, 0

2 1.8 1.6 1.4 1.2 1 -1.5

53. (a) Let f (x) = x2 +

0.5

1.5

55. (a) (−2.2, 4), (2, 1.2), (4.2, 3) (b) f exists everywhere, so the critical numbers are when f = 0, i.e. when x = ±2 or r(x) = 0, so x ≈ −5.1, −2, 0.2, 2. At x = −5.1 f changes sign from − to +, so minimum; at x = −2 f changes sign from + to −, so maximum; at x = 0.2 f doesn’t change sign, so neither; at x = 2 f changes sign from − to +, so minimum. Finally, f (1) = (12 − 4)r (1) + 2r(1) ≈ −3(0.6) + 2(0.3) = −1.2.

uh a

56. g (x) exists everywhere, so the critical points are the points when g (x) = 0, or r(x) = x, so r(x) crosses the line y = x. From the graph it appears that this happens precisely when x = 0.

57. f (x) = 3ax2 + 2bx + c and f (x) has roots at x = 0, 1, so f (x) must be of the form f (x) = 3ax(x − 1); thus c = 0 and 2b = −3a, b = −3a/2. f (x) = 6ax + 2b = 6ax − 3a, so f (0) > 0 and f (1) < 0 provided a < 0. Finally f (0) = d, so d = 0; and f (1) = a + b + c + d = a + b = −a/2 so a = −2. Thus f (x) = −2x3 + 3x2 .

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Chapter 5

58. (a) one relative maximum, located at x = n (b) f (x) = cxn−1 (−x + n)e−x = 0 at x = n. Since f (x) > 0 for x < n and f (x) < 0 for x > n it’s a maximum.

0.3

(b)

1 2c

( µ,

)

1 2c

59. (a) f (x) = −xf (x). Since f (x) is always positive, f (x) = 0 at x = 0, f (x) > 0 for x < 0 and f (x) < 0 for x > 0, so x = 0 is a maximum.

us uf

164

(b)

)

(

(c)

(

f (x0 ) is a relative maximum.

EXERCISE SET 5.3

x 1 -4

(1, -4)

uh a

1. y = x2 − 2x − 3; y = 2(x − 1); y = 2

)

f (x0 ) is a relative minimum.

f (x0 ) is not an extreme value.

61. (a)

60. (a) Because h and g have relative maxima at x0 , h(x) ≤ h(x0 ) for all x in I1 and g(x) ≤ g(x0 ) for all x in I2 , where I1 and I2 are open intervals containing x0 . If x is in both I1 and I2 then both inequalities are true and by addition so is h(x) + g(x) ≤ h(x0 ) + g(x0 ) which shows that h + g has a relative maximum at x0 . (b) By counterexample; both h(x) = −x2 and g(x) = −2x2 have relative maxima at x = 0 but h(x) − g(x) = x2 has a relative minimum at x = 0 so in general h − g does not necessarily have a relative maximum at x0 .

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Exercise Set 5.3

(

4. y = 2x3 − 3x2 + 12x + 9; y = 6(x2 − x + 2); y = 12(x − 1/2) y

y 1, 5 2 4

)

3. y = x3 − 3x + 1; y = 3(x2 − 1); y = 6x

(-1, 3)

1 (0, 1)

(12, 292)

(1, -1)

7. y = x3 (3x2 − 5); y = 15x2 (x2 − 1); y = 30x(2x2 − 1)

6. y = x4 − 2x2 − 12; y = 4x(x2 − 1); y = 12(x2 − 1/3)

5. y = x4 + 2x3 − 1; y = 4x2 (x + 3/2); y = 12x(x + 1)

(

-1 ,7 3 3 8

)

(-1, -13)

(1, -13)

(- 13 , - 1139 )

( 13 , - 1139 )

(

9. y = x(x − 1)3 ; y = (4x − 1)(x − 1)2 ; y = 6(2x − 1)(x − 1)

(1, -2)

(

1 , -7 3 8 3

)

10. y = x4 (x + 5); y = 5x3 (x + 4); y = 20x2 (x + 3)

8. y = 3x3 (x + 4/3); y = 12x2 (x + 1); y = 36x(x + 2/3)

(0, 0)

(0, -12)

(-1, -2) 3 43 ,2 16

)

(0, -1)

(-1, 2)

us uf

2. y = 1 + x − x2 ; y = −2(x − 1/2); y = −2

165

y (-4, 256)

300

0.1

x (1, 0)

(0, 0)

2 16 - ,3 27

(

)

-0.1

(0, 0) (-3, 162)

y y=2 x

x=3

uh a

11. y = 2x/(x − 3); y = −6/(x − 3)2 ; y = 12/(x − 3)3

)

27 ( 14 , - 256 )

(-1, -1)

(

1 1 ,2 16

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Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Chapter 5

y = − y =

x2 ; −1 2x y = − 2 ; (x − 1)2

x ; −1

13. y =

x2 + 1 ; (x2 − 1)2

2x(x2 + 3) (x2 − 1)3

y =

12. y =

us uf

166

2(3x2 + 1) (x2 − 1)3 y

x = -1

y=1

(0, 0)

x (0, 0)

x2 − 1 ; x2 + 1 4x ; y = 2 (x + 1)2

14. y =

x=1

y = -1

x=1

y=1

4(1 − 3x2 ) y = (x2 + 1)3

(- 13 , - 12 )

( 13 , - 12 )

x3 − 1 1 = ; x x

2x3 + 1 , x2 y = 0 when 3 1 ≈ −0.8; x=− 2

17. y =

x3 − 1 ; x3 + 1

y =

6x2 ; (x3 + 1)2

y =

12x(1 − 2x3 ) (x3 + 1)3 y

y=2 x

2(x3 − 1) y = x3

y =

2x2 − 1 ; x2 2 y = 3 ; x 6 y = − 4 x

16. y =

15. y = x2 −

(0, -1)

y=1

y ≈(-0.8, 1.9)

x = -1

(1/ 2 , -1/3) (0, -1)

uh a

(1, 0)

x 3

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Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Exercise Set 5.3

y =

19. y =

16(3x2 + 4) (4 − x2 )3

x−1 ; x2 − 4 x2 − 2x + 4 (x2 − 4)2

y = 2

x3 − 3x2 + 12x − 4 (x2 − 4)3

y x=2

x = -2

y = −

us uf

8 ; 4 − x2 16x y = ; (4 − x2 )2

18. y =

(0, 2) x

(0.36, 0.16)

x =2

1 ; x−2 −1 y = (x − 2)2

x+3 ; x2 − 4

21. y =

(x2 + 6x + 4) (x2 − 4)2

y = 2

x3 + 9x2 + 12x + 12 (x2 − 4)3

y = −

y –3 – ﬁ x

-3 + ﬁ

-52/3-51/3-3

y = −

x(x + 2) + x + 1)2

(x2

x3 + 3x2 − 1 (x2 + x + 1)3

y = 2

x2 − 1 x+1 = 2 ; 3 x −1 x +x+1

22. y =

uh a

20. y =

(x − 1)2 ; x2 2(x − 1) ; y = x3 2(3 − 2x) y = x4

23. y =

y (0, 1) (0.53, 0.84)

(–0.65, 0.45) x

(–2, – 1 ) 3

(–2.88, –0.29) –1

y =1

–4

167

x (1, 0)

( 32 , 19 )

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Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Chapter 5

4 4 − ; x x2 4(x + 2) ; y = x3 8(x + 3) y = − x4

24. y = 3 −

y -3, 35 9

)

(

(-2, 4)

us uf

168

y=3

x−1 ; x4 3x − 4 y = − ; x5 3x − 5 y = 4 x6

26. y = 2 + y =

3(1 − x2 ) ; x4

y =

6(x2 − 2) x5

(

)

2554 625

(

2, 2 +

5 4

y=2

2 , 2 – 54 2)

(−

–2

(b) I

(d) V

(–1, 0)

(e) IV

(f ) II

27. (a) VI

(1, 4)

( 43 , 1051 256 ) 5 3

1 3 − 3; x x

25. y = 4 +

28. (a) When n is even the function is deﬁned only for x ≥ 0; as n increases the graph approaches the line y = 1 for x > 0.

uh a

(b) When n is odd the graph is symmetric with respect to the origin; as n increases the graph approaches the line y = 1 for x > 0 and the line y = −1 for x < 0.

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Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Exercise Set 5.3

y =

x2 − 4;

2x ; 3(x2 − 4)2/3

y = −

30. y =

2(3x2 + 4) 9(x2 − 4)5/3

y 3

3 x

(-2, 0)

(2, 0)

-1

us uf

√

x2 − 1; x y = √ ; 2 x −1 1 y = − 2 (x − 1)3/2

29. y =

-2

(0, -2)

32. y = 4x − 3x4/3 ;

y = 2 + 2x−1/3 ;

y = 4 − 4x1/3 ;

2 y = − x−4/3 3

4 y = − x−2/3 3

33. y = x(3 − x)1/2 ; 3(2 − x) ; y = √ 2 3−x

31. y = 2x + 3x2/3 ;

y =

y 3

3(x − 4) 4(3 − x)3/2 y

(1, 1)

(2, 2) x

x (0, 0)

34. y = x1/3 (4 − x); 4(1 − x) ; 3x2/3 4(x + 2) y = − 9x5/3 y

y =

(1, 3)

(-2, -6

)

√ 8( x − 1) ; x √ 4(2 − x) ; y = x2 √ 2(3 x − 8) y = x3

35. y =

169

y (4, 2)

(649 , 158 ) x 15

uh a

-10

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Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Chapter 5

√ 1+ x √ ; 1− x

y x=1

1 √ ; y = √ 2 x(1− x) √ 3 x−1 √ y = 3/2 2x (1 − x)3

us uf

36. y =

( 19 , 2 ) x y = -1

-2

170

38. y = x − cos x; y = 1 + sin x; y = 0 when x = −π/2 + 2nπ; y = cos x; y = 0 when x = π/2 + nπ n = 0, ±1, ±2, . . .

37. y = x + sin x; y = 1 + cos x, y = 0 when x = π + 2nπ; y = − sin x; y = 0 when x = nπ n = 0, ±1, ±2, . . . y

√ 40. y = 3√cos x + sin x; y = − 3 sin x + cos x; x = π/6 + nπ; y = 0 when √ y = − 3 cos x − sin x; y = 0 when x = 2π/3 + nπ

39. y = sin x + cos x; y = cos x − sin x; y = 0 when x = π/4 + nπ; y = − sin x − cos x; y = 0 when x = 3π/4 + nπ

y 2

o -2

-o

uh a

41. y = sin2 x, 0 ≤ x ≤ 2π; y = 2 sin x cos x = sin 2x; y = 2 cos 2x

-2

42. y = x tan x, −π/2 < x < π/2; y = x sec2 x + tan x; y = 0 when x = 0; y = 2 sec2 x(x tan x + 1), which is always positive for −π/2 < x < π/2 y

x o

x -6

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Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Exercise Set 5.3

lim xex = +∞, lim xex = 0

x→+∞

x→−∞

(b) y = xex ; y = (x + 1)ex ; y = (x + 2)ex

-3

(-2, -0.27) (-1, -0.37)

x -1

lim xe−2x = 0, lim xe−2x = −∞ x→−∞

x→+∞

−2x

(b) y = xe

1 ; y = −2 x − 2

y 0.3

−2x

(0.5, 0.18) (1, 0.14)

44. (a)

1 -5

us uf

43. (a)

171

−2x

; y = 4(x − 1)e

0.1

-3

-0.3

x2 x2 = 0, lim = +∞ x→+∞ e2x x→−∞ e2x lim

45. (a)

0.3

y = 2x(1 − x)e−2x ; y = 2(2x2 − 4x + 1)e−2x ;

lim x2 e2x = +∞, lim x2 e2x = 0.

x→+∞

x→−∞

(b) y = x2 e2x ;

0.3

y = 0 if 2x2 + 4x + 1 = 0, when √ √ −4 ± 16 − 8 = −1 ± 2/2 ≈ −0.29, −1.71 x= 4

(-0.29, 0.05)

-3 -2 -1

lim f (x) = +∞, lim f (x) = −∞

x→+∞

100

(0,0)

-2

2 -100

lim f (x) = 1

x→±∞

uh a

(0, 0)

x→−∞

1 2

(b) f (x) = 2x−3 e−1/x so f (x) < 0 for x < 0 and f (x) > 0 for x > 0. Set u = x2 and use the given result to ﬁnd lim f (x) = 0, so (by the First Derivative Test) f (x) x→0

y = 2(2x2 + 4x + 1)e2x ;

(b) y = xex ; 2 y = (1 + 2x2 )ex ; 2 y = 2x(3 + 2x2 )ex no relative extrema, inﬂection point at (0, 0) 48. (a)

0.2

(-1, 0.14) (-1.71, 0.10)

y = 2x(x + 1)e ;

47. (a)

46. (a)

(1, 0.14) (1.71, 0.10) x 1 2 3 (0, 0) (0.29, 0.05)

y = 0 if 2x2 − 4x + 1 = 0, when √ √ 4 ± 16 − 8 = 1 ± 2/2 ≈ 0.29, 1.71 x= 4

(b) y = x2 /e2x = x2 e−2x ;

−6 −1/x , has a minimum at x = 0. f (x) = (−6x−4 + 4x )e so f (x) has points of inﬂection at x = ± 2/3.

0.8 0.4

(-√2/3, e-3/2)

(√2/3, e-3/2) x

-10 -5

5 10 (0, 0)

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172

49.

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Chapter 5

lim f (x) = +∞, lim f (x) = 0

x→+∞

x→−∞

x−1 x − 2x + 2 f (x) = ex 2 , f (x) = ex x x3 critical point at x = 1; relative minimum at x = 1 no points of inﬂection vertical asymptote x = 0, horizontal asymptote y = 0 for x → −∞ 2

i x

–4

lim f (x) = 0, lim f (x) = −∞

x→−∞

x→+∞

–2

f (x) = (1 − x)e−x , f (x) = (x − 2)e−x critical point at x = 1; relative maximum at x = 1 point of inﬂection at x = 2 horizontal asymptote y = 0 as x → +∞

(1, e1)

0.2

(2, e22)

us uf

(1, e)

50.

-0.8

lim f (x) = 0, lim f (x) = +∞

x→+∞

1.8

x→−∞ 1−x

lim f (x) = +∞, lim f (x) = 0

x→+∞

x→−∞ x−1

0.6

(0.59, 0.52) x 1

(0, 0)

x→0

lim y = +∞

x→+∞

ln x 1/x = lim = 0; 1/x x→0+ −1/x2

(b) y = x ln x, y = 1 + ln x, y = 1/x, y = 0 when x = e−1 lim y = lim+

x→0+

uh a

54. (a)

x→0

lim y = +∞

(b) y = x2 ln x, y = x(1 + 2 ln x), y = 3 + 2 ln x, y = 0 if x = e−1/2 , y = 0 if x = e−3/2 , lim y = 0

y 0.8 0.4 (0, 0) -4 (–4.7, –0.35)

-2 -0.4 (–3, –0.49)

x 1 (–1.27, –0.21)

x 1 (e-1, -e-1)

ln x 1/x = lim+ = 0, 1/x2 x→0 −2/x3

x→+∞

x−1

lim y = lim+ x ln x = lim+

x→0+

(3.41, 1.04)

f (x) = x (3 + x)e , f (x) = x(x + 6x + 6)e critical points at x = −3, 0; relative minimum at x = −3 √ points of inﬂection at x = 0, −3 ± 3 ≈ 0, −4.7, −1.27 horizontal asymptote y = 0 as x → −∞

53. (a)

(2, 4e )

52.

f (x) = x(2 − x)e , f (x) = (x2 − 4x + 2)e1−x critical points at x = 0, 2; relative minimum at x = 0, relative maximum at x = 2 √ points of inﬂection at x = 2 ± 2 horizontal asymptote y = 0 as x → +∞

51.

y 0.2 0.1

(e-3/2, - 32 e-3)-0.1 -0.2

x 1

(e-1/2, -12 e-1)

x→0+

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Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Exercise Set 5.3

ln x = −∞; x2 ln x 1/x =0 lim y = lim = lim x→+∞ x→+∞ x2 x→+∞ 2x

(e1/2, 12 e-1) (e5/6, 56 e-5/3)

0.4 0.3 0.2 0.1

x→0+

ln x 1 − 2 ln x (b) y = 2 , y = , x x3 6 ln x − 5 y = , x4 y = 0 if x = e1/2 , y = 0 if x = e5/6

x→0+

x 1

-0.1 -0.2 -0.3 -0.4

us uf

lim y = lim

55. (a)

173

√ √ 56. (a) Let u = 1/x, lim+ (ln x)/ x = lim − u ln u = −∞ by inspection, u→+∞ x→0 √ lim (ln x)/ x = 0, by the rule given. y 2 (e , 2/e)

x→+∞

(e8/3, 83 e-4/3)

ln x 2 − ln x (b) y = √ , y = x 2x3/2 −8 + 3 ln x y = 4x5/2 y = 0 if x = e2 , y = 0 if x = e8/3

57. (a)

0.5

an ss

(2e1 , – 8e3 ) 3/2

(21 e, – 8e1 ) √

lim f (x) = +∞; lim f (x) = 0

x→+∞

x→0

2 1

(–1, ln 2)

x→+∞

uh a

curve crosses x-axis at x = 0, 1, −1

-2

(1, ln 2) x

–2

lim y = −∞, lim y = +∞;

x→−∞

(b)

(0, 0)

lim y = +∞;

x→±∞

curve never crosses x-axis y

y 4

0.2

-1

x 1 2

(b) y = ln(x2 + 1), y = 2x/(x2 + 1) x2 − 1 y = −2 2 (x + 1)2 y = 0 if x = 0 y = 0 if x = ±1

x→+∞

(b) y = x2 ln 2x, y = 2x ln 2x + x y = 2 ln 2x + 3 √ y = 0 if x = 1/(2 e), y = 0 if x = 1/(2e3/2 )

59. (a)

lim x2 ln x = 0 by the rule given, lim x2 ln x = +∞ by inspection, and f (x) not deﬁned

x→0+

for x < 0

58. (a)

-1

az 2

x 1

0.1

-2 -4 -6

x -1

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Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Chapter 5

lim y = −∞, lim y = +∞;

x→−∞

(d)

x→+∞

lim y = +∞;

x→±∞

curve crosses x-axis at x = −1

curve crosses x-axis at x = 0, 1 y

y 0.4

0.4

0.2

-1

-0.2

x a

(c)

61. (a) horizontal asymptote y = 3 as x → ±∞, vertical asymptotes of x = ±2

(b) horizontal asymptote of y = 1 as x → ±∞, vertical asymptotes at x = ±1

y 10

uh a

(b)

-1

60. (a)

(c)

us uf

174

5 x x

-5

5 -10

-5

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Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Exercise Set 5.3

(d) horizontal asymptote of y = 1 as x → ±∞, vertical asymptote at x = −1, 2

y 10

-5

-10 -10

x b

a+b means f 2

a+b a+b +x =f − x for any x. 2 2

(b) Symmetric about the line x =

-5

us uf

62. (a)

175

a−b a+b a+b a−b +x−a = x− and +x−b = x+ , and the same equations 2 2 2 2 are true with x replaced by −x. Hence 2 a+b a−b a−b a+b a−b +x−a +x−b = x− x+ = x2 − 2 2 2 2 2

Note that

The right hand side remains the same if we replace x with −x, and so the same is true of the left hand side, and the same is therefore true of the reciprocal of the left hand side. But a+b + x . Since this quantity remains the reciprocal of the left hand side is equal to f 2 a+b unchanged if we replace x with −x, the condition of symmetry about the line x = has 2 been met. 63. (a)

uh a

0.4

-0.5

(b) y = (1 − bx)e−bx , y = b2 (x − 2/b)e−bx ; relative maximum at x = 1/b, y = 1/be; point of inﬂection at x = 2/b, y = 2/be2 . Increasing b moves the relative maximum and the point of inﬂection to the left and down, i.e. towards the origin.

-0.2

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176

Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Chapter 5

64. (a)

(b) y = −2bxe−bx , 2 y = 2b(−1 + 2bx2 )e−bx ; relative maximum at x = 0, y = 1; points √ of inﬂection at x = ± 1/2b, y = 1/ e. Increasing b moves the points of inﬂection towards the y-axis; the relative maximum doesn’t move.

-2

2 0

65. (a) The oscillations of ex cos x about zero increase as x → +∞ so the limit does not exist, and lim ex cos x = 0.

(b)

6 (0,1) 4

(1.52, 0.22) x 1 2

-2 -1

x→−∞

us uf

-2

(c) The curve y = eax cos bx oscillates between y = eax and y = −eax . The frequency of oscillation increases when b increases. a=3

b=3 5

b=1

x→±∞

b=2

b=1

P (x) − (ax + b) = Q(x)

R(x) = 0 because the degree of R(x) is less than x→±∞ Q(x) lim

the degree of Q(x).

x2 − 2 2 = x − so x x y = x is an oblique asymptote;

67. y =

x2 + 2 , x2 4 y = − 3 x

lim

0.5 1

-5 a=1

-1

66.

a=2 a=1

y =

y 4

5 x2 − 2x − 3 =x−4+ so x+2 x+2 y = x − 4 is an oblique asymptote;

68. y =

x2 + 4x − 1 10 ,y = 2 (x + 2) (x + 2)3

x = -2 x -10 y =x-4

10 ≈(0.24, -1.52) ≈(-4.24, -10.48)

uh a

y =

y=x

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Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Exercise Set 5.3

(x − 2)3 12x − 8 =x−6+ so 2 x x2 y = x − 6 is an oblique asymptote;

69. y =

y (2, 0)

(x − 2)2 (x + 4) y = , x3 24(x − 2) y = x4

us uf

x -10

y =x-6

(-4, -13.5)

4 − x3 , x2

x3 + 8 , x3 24 y = 4 x y = −

(-2, 3)

70. y =

177

y = -x

(x + 1)(x2 − x + 2) , x3 2(x + 3) y = − x4

y =

1 (x − 1)(x + 1)2 1 , − 2 = x x x2 y = x + 1 is an oblique asymptote;

71. y = x + 1 −

y = x+1

(-1, 0)

(-3, -169 )

lim [f (x) − x2 ] = lim (1/x) = 0

x→±∞

73.

72. The oblique asymptote is y = 2x so (2x3 − 3x + 4)/x2 = 2x, −3x + 4 = 0, x = 4/3.

1 1 2x3 − 1 x3 + 1 , = , y = 2x − 2 = x x x x2 √ 2 2(x3 + 1) , y = 0 when x = 1/ 3 2 ≈ 0.8, y = 2 + 3 = x x3 √ y = 3 3 2/2 ≈ 1.9; y = 0 when x = −1, y = 0

uh a

y = x2 +

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y = x2 x 3

Chapter 5

lim [f (x) − (3 − x2 )] = lim (2/x) = 0

x→±∞

y = 3 − x2 +

2 + 3x − x3 2 2 2(x3 + 1) = , y = −2x − 2 = − , x x x x2

74.

Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

1000

us uf

178

4 2(x − 2) =− , y = 0 when x = −1, y = 0; x3 x3 √ y = 0 when x = 3 2 ≈ 1.3, y = 3 3

y = −2 +

75. Let y be the length of the other side of the rectangle, then L = 2x+2y and xy = 400 so y = 400/x and hence L = 2x + 800/x. L = 2x is an oblique asymptote (see Exercise 66) 2(x2 + 400) 800 = , x x

100

L = 2x +

2(x2 − 400) 800 = , 2 x x2 1600 L = 3 , x L = 0 when x = 20, L = 80 L = 2 −

76. Let y be the height of the box, then S = x2 + 4xy and x2 y = 500 so y = 500/x2 and hence S = x2 + 2000/x. The graph approaches the curve S = x2 asymptotically (see Exercise 73)

S = 2x −

2000 2(x3 − 1000) = , 2 x x2

x + 2000 2000 = , x x

1000

S = x2 +

x 30

4000 2(x3 + 2000) = , 3 x x3 S = 0 when x = 10, S = 300

S = 2 +

77. y = 0.1x4 (6x − 5); critical numbers: x = 0, x = 5/6; relative minimum at x = 5/6, y ≈ −6.7 × 10−3

uh a

78. y = 0.1x4 (x + 1)(7x + 5); critical numbers: x = 0, x = −1, x = −5/7, relative maximum at x = −1, y = 0; relative minimum at x = −5/7, y ≈ −1.5 × 10−3

0.01

-1

0.001

x 1

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Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Exercise Set 5.4

179

(b) positive, positive, speeding up

2. (a) positive, slowing down (c) positive, speeding up

(b) negative, slowing down

left because v = ds/dt < 0 at t0 negative because a = d2 s/dt2 and the curve is concave down at t0 (d2 s/dt2 < 0) speeding up because v and a have the same sign v < 0 and a > 0 at t1 so the particle is slowing down because v and a have opposite signs.

4. (a) III

3. (a) (b) (c) (d)

us uf

1. (a) positive, negative, slowing down (c) negative, positive, slowing down

EXERCISE SET 5.4

(b) I

5. s (m)

t (s)

6. (a) when s ≥ 0, so 0 < t < 2 and 4 < t ≤ 8 (c) when s is decreasing, so 0 ≤ t < 3

|v|

10 5 0

t 1

(b) when the slope is zero, at t = 3

-5

t 6

–15

-10

uh a

(b)

8. (a) v ≈ (30 − 10)/(15 − 10) = 20/5 = 4 m/s a

25 (1)

25 (2)

9. (a) At 60 mi/h the slope of the estimated tangent line is about 4.6 mi/h/s. Use 1 mi = 5, 280 ft and 1 h = 3600 s to get a = dv/dt ≈ 4.6(5,280)/(3600) ≈ 6.7 ft/s2 . (b) The slope of the tangent to the curve is maximum at t = 0 s.

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Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Chapter 5

10. (a)

t 1 2 3 4 5 s 0.71 1.00 0.71 0.00 −0.71 v 0.56 0.00 −0.56 −0.79 −0.56 a −0.44 −0.62 −0.44 0.00 0.44

us uf

180

(b) to the right at t = 1, stopped at t = 2, otherwise to the left (c) speeding up at t = 3; slowing down at t = 1, 5; neither at t = 2, 4

v(t) = 3t2 − 12t, a(t) = 6t − 12 s(1) = −5 ft, v(1) = −9 ft/s, speed = 9 ft/s, a(1) = −6 ft/s2 v = 0 at t = 0, 4 for t ≥ 0, v(t) changes sign at t = 4, and a(t) changes sign at t = 2; so the particle is speeding up for 0 < t < 2 and 4 < t and is slowing down for 2 < t < 4 (e) total distance = |s(4) − s(0)| + |s(5) − s(4)| = | − 32 − 0| + | − 25 − (−32)| = 39 ft

v(t) = 4t3 − 4, a(t) = 12t2 s(1) = −1 ft, v(1) = 0 ft/s, speed = 0 ft/s, a(1) = 12 ft/s2 v = 0 at t = 1 speeding up for t > 1, slowing down for 0 < t < 1 total distance = |s(1) − s(0)| + |s(5) − s(1)| = | − 1 − 2| + |607 − (−1)| = 611 ft

12. (a) (b) (c) (d) (e)

11. (a) (b) (c) (d)

v(t) = −(3π/2) sin(πt/2), a(t) = −(3π 2 /4) cos(πt/2) s(1) = 0 ft, v(1) = −3π/2 ft/s, speed = 3π/2 ft/s, a(1) = 0 ft/s2 v = 0 at t = 0, 2, 4 v changes sign at t = 0, 2, 4 and a changes sign at t = 1, 3, 5, so the particle is speeding up for 0 < t < 1, 2 < t < 3 and 4 < t < 5, and it is slowing down for 1 < t < 2 and 3 < t < 4 (e) total distance = |s(2) − s(0)| + |s(4) − s(2)| + |s(5) − s(4)| = | − 3 − 3| + |3 − (−3)| + |0 − 3| = 15 ft

13. (a) (b) (c) (d)

4 − t2 2t(t2 − 12) , a(t) = 2 2 (t + 4) (t2 + 4)3 s(1) = 1/5 ft, v(1) = 3/25 ft/s, speed = 3/25 ft/s, a(1) = −22/125 ft/s2 v = 0 at t = 2 √ √ a changes sign at t = 2 3, so √ the particle is speeding up for 2 < t < 2 3 and it is slowing down for 0 < t < 2 and for 2 3 < t 1 5 1 19 total distance = |s(2) − s(0)| + |s(5) − s(2)| = − 0 + − = ft 4 29 4 58

(e)

uh a

0.25

2t(t2 − 15) 5 − t2 , a(t) = 2 2 (t + 5) (t2 + 5)3

0.01 0

0.2

15. v(t) =

(b) (c) (d)

14. (a) v(t) =

20 -0.05

s(t)

(a) v = 0 at t =

√

-0.15

v(t)

(b) s =

√

5/10 at t =

a(t)

√

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Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Exercise Set 5.4

181

√ √ √ (c) a changes sign √ at t =√ 15, so the particle is speeding up for 5 < t < 15 and slowing down for 0 < t < 5 and 15 < t

us uf

16. v(t) = (1 − t)e−t , a(t) = (t − 2)e−t 0.5 0

2 0

-0.2

s(t)

-2

v(t)

a(t)

(a) v = 0 at t = 1 (b) s = 1/e at t = 1 (c) a changes sign at t = 2, so the particle is speeding up for 1 < t < 2 and slowing down for 0 < t < 1 and 2 < t 17. s = −3t + 2 v = −3 a=0

Constant speed

t=0

18. s = t3 − 6t2 + 9t + 1 v = 3(t − 1)(t − 3) a = 6(t − 2)

Speeding up

t=2

(Stopped) t = 3

t = 1 (Stopped)

t=0

19. s = t3 − 9t2 + 24t v = 3(t − 2)(t − 4) a = 6(t − 3)

Slowing down

Speeding up Slowing down (Stopped) t = 4 t=3

t = 2 (Stopped) s 16 18 20

t=0 0

20. s = t +

9 t+1

Slowing down

(t + 4)(t − 2) v= (t + 1)2

Speeding up (Stopped) t = 2

s 5 Slowing down 9

18 (t + 1)3

cos t, 0 ≤ t ≤ 2π 1, t > 2π

uh a

21. s =

− sin t, 0 ≤ t ≤ 2π 0, t > 2π − cos t, 0 ≤ t < 2π 0, t > 2π

Slowing down t=c -1

t = 3c/2 t = c/2 0

(Stopped permanently) t = 2c t=0 s 1 Speeding up

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Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Chapter 5

22. v(t) =

5t2 − 6t + 2 √ is always positive, t

Slowing down

√ 15t2 − 6t − 2 3 + 39 a(t) = has a positive root at t = 15 2t3/2

Speeding up s

us uf

3 + 39 15

182

23. (a) v = 10t − 22, speed = |v| = |10t − 22|. d|v|/dt does not exist at t = 2.2 which is the only critical point. If t = 1, 2.2, 3 then |v| = 12, 0, 8. The maximum speed is 12 ft/s. (b) the distance from the origin is |s| = |5t2 − 22t| = |t(5t − 22)|, but t(5t − 22) < 0 for 1 ≤ t ≤ 3 so |s| = −(5t2 − 22t) = 22t − 5t2 , d|s|/dt = 22 − 10t, thus the only critical point is t = 2.2. d2 |s|/dt2 < 0 so the particle is farthest from the origin when t = 2.2. Its position is s = 5(2.2)2 − 22(2.2) = −24.2. 200t 200t d|v| 600(4 − t2 ) , speed = |v| = for t ≥ 0. = 0 when t = 2, which = dt (t2 + 12)3 (t2 + 12)2 (t2 + 12)2 is the only critical point in (0, +∞). By the ﬁrst derivative test there is a relative maximum, and hence an absolute maximum, at t = 2. The maximum speed is 25/16 ft/s to the left.

24. v = −

25. s(t) = s0 − 12 gt2 = s0 − 4.9t2 m, v = −9.8t m/s, a = −9.8 m/s2 (a) |s(1.5) − s(0)| = 11.025 m (b) v(1.5) = −14.7 m/s (c) |v(t)| = 12 when t = 12/9.8 = 1.2245 s (d) s(t) − s0 = −100 when 4.9t2 = 100, t = 4.5175 s

√ 800 =5 2 = 800 − 16t ft, s(t) = 0 when t = 16 √ √ (b) v(t) = −32t and v(5 2) = −160 2 ≈ 226.27 ft/s = 154.28 mi/h

26. (a) s(t) = s0 −

1 2 2 gt

27. s(t) = s0 + v0 t − 12 gt2 = 60t − 4.9t2 m and v(t) = v0 − gt = 60 − 9.8t m/s (a) v(t) = 0 when t = 60/9.8 ≈ 6.12 s (b) s(60/9.8) ≈ 183.67 m (c) another 6.12 s; solve for t in s(t) = 0 to get this result, or use the symmetry of the parabola s = 60t − 4.9t2 about the line t = 6.12 in the t-s plane (d) also 60 m/s, as seen from the symmetry of the parabola (or compute v(6.12))

28. (a) they are the same (b) s(t) = v0 t − 12 gt2 and v(t) = v0 − gt; s(t) = 0 when t = 0, 2v0 /g; v(0) = v0 and v(2v0 /g) = v0 − g(2v0 /g) = −v0 so the speed is the same at launch (t = 0) and at return (t = 2v0 /g).

29. If g = 32 ft/s2 , s0 = 7 and v0 is unknown, then s(t) = 7 + v0 t − 16t2 and v(t) = v0 − 32t; s = smax when v = 0, or t = v0 /32; and smax = 208 yields √ 208 = s(v0 /32) = 7 + v0 (v0 /32) − 16(v0 /32)2 = 7 + v02 /64, so v0 = 8 201 ≈ 113.42 ft/s.

uh a

30. (a) Use (6) and then (5) to get v 2 = v02 − 2v0 gt + g 2 t2 = v02 − 2g(v0 t − 12 gt2 ) = v02 − 2g(s − s0 ). (b) Add v0 to both sides of (6): 2v0 − gt = v0 + v, v0 − 12 gt = 21 (v0 + v); from (5) s = s0 + t(v0 − 12 gt) = s0 + 12 (v0 + v)t (c) Add v to both sides of (6): 2v + gt = v0 + v, v + 12 gt = 12 (v0 + v); from Part (b), s = s0 + 12 (v0 + v)t = s0 + vt + 12 gt2

31. v0 = 0 and g = 9.8, so v 2 = −19.6(s − s0 ); since v = 24 when s = 0 it follows that 19.6s0 = 242 or s0 = 29.39 m.

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Exercise Set 5.4

183

32. s = 1000 + vt + 12 (32)t2 = 1000 + vt + 16t2 ; s = 0 when t = 5, so v = −(1000 + 16 · 52 )/5 = −280 ft/s.

us uf

33. (a) s = smax when v = 0, so 0 = v02 − 2g(smax − s0 ), smax = v02 /2g + s0 . (b) s0 = 7,√smax = 208, g = 32 and v0 is unknown, so from Part (a) v02 = 2g(208 − 7) = 64 · 201, v0 = 8 201 ≈ 113.42 ft/s.

(b) v = √

1.5

√ 2t 2 , lim v = √ = 2 2 2t2 + 1 t→+∞

35. (a)

34. s = t3 − 6t2 + 1, v = 3t2 − 12t, a = 6t − 12. (a) a = 0 when t = 2; s = −15, v = −12. (b) v = 0 when 3t2 − 12t = 3t(t − 4) = 0, t = 0 or t = 4. If t = 0, then s = 1 and a = −12; if t = 4, then s = −31 and a = 12.

dv dv ds dv ds = =v because v = dt ds dt ds dt 9 3 3 3 dv (b) v = √ = − 2 ; a = − 3 = −9/500 = ; 2s ds 2s 4s 2 3t + 7

36. (a) a =

37. (a) s1 = s2 if they collide, so 12 t2 − t + 3 = − 14 t2 + t + 1, 34 t2 − 2t + 2 = 0 which has no real solution. (b) Find the minimum value of D = |s1 − s2 | = 34 t2 − 2t + 2 . From Part (a), 34 t2 − 2t + 2

is never zero, and for t = 0 it is positive, hence it is always positive, so D =

3 2 4t

− 2t + 2.

dD d2 D = 32 t − 2 = 0 when t = 43 . > 0 so D is minimum when t = 43 , D = 23 . dt dt2 1 (c) v1 = t − 1, v2 = − t + 1. v1 < 0 if 0 ≤ t < 1, v1 > 0 if t > 1; v2 < 0 if t > 2, v2 > 0 if 2 0 ≤ t < 2. They are moving in opposite directions during the intervals 0 ≤ t < 1 and t > 2.

38. (a) sA − sB = 20 − 0 = 20 ft (b) sA = sB , 15t2 + 10t + 20 = 5t2 + 40t, 10t2 − 30t + 20 = 0, (t − 2)(t − 1) = 0, t = 1 or t = 2 s. (c) vA = vB , 30t+10 = 10t+40, 20t = 30, t = 3/2 s. When t = 3/2, sA = 275/4 and sB = 285/4 so car B is ahead of car A.

uh a

39. r(t) = v 2 (t), r (t) = 2v(t)v (t)/[2 v(t)] = v(t)a(t)/|v(t)| so r (t) > 0 (speed is increasing) if v and a have the same sign, and r (t) < 0 (speed is decreasing) if v and a have opposite signs. If v(t) > 0 then r(t) = v(t) and r (t) = a(t), so if a(t) > 0 then the particle is speeding up and a and v have the same sign; if a(t) < 0, then the particle is slowing down, and a and v have opposite signs. If v(t) < 0 then r(t) = −v(t), r (t) = −a(t), and if a(t) > 0 then the particle is speeding up and a and v have opposite signs; if a(t) < 0 then the particle is slowing down and a and v have the same sign.

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184

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Chapter 5

EXERCISE SET 5.5

us uf

1. relative maxima at x = 2, 6; absolute maximum at x = 6; relative and absolute minimum at x = 4

2. relative maximum at x = 3; absolute maximum at x = 7; relative minima at x = 1, 5; absolute minima at x = 1, 5 y

3. (a)

(b)

x 10 y

(c)

x 5

4. (a)

(b)

-5

(c)

5. f (x) = 8x − 4, f (x) = 0 when x = 1/2; f (0) = 1, f (1/2) = 0, f (1) = 1 so the maximum value is 1 at x = 0, 1 and the minimum value is 0 at x = 1/2.

6. f (x) = 8 − 2x, f (x) = 0 when x = 4; f (0) = 0, f (4) = 16, f (6) = 12 so the maximum value is 16 at x = 4 and the minimum value is 0 at x = 0.

uh a

7. f (x) = 3(x − 1)2 , f (x) = 0 when x = 1; f (0) = −1, f (1) = 0, f (4) = 27 so the maximum value is 27 at x = 4 and the minimum value is −1 at x = 0.

8. f (x) = 6x2 − 6x − 12 = 6(x + 1)(x − 2), f (x) = 0 when x = −1, 2; f (−2) = −4, f (−1) = 7, f (2) = −20, f (3) = −9 so the maximum value is 7 at x = −1 and the minimum value is −20 at x = 2. √ √ 2 3/2 9. f (x) √ = −3/ 5, f (1) = 3/ 5 so the maximum value is √ = 3/(4x + 1) , no critical points; f (−1) 3/ 5 at x = 1 and the minimum value is −3/ 5 at x = −1.

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Exercise Set 5.5

185

2(2x + 1) , f (x) = 0 when x = −1/2 and f (x) does not exist when x = −1, 0; 3(x2 + x)1/3 f (−2) = 22/3 , f (−1) = 0, f (−1/2) = 4−2/3 , f (0) = 0, f (3) = 122/3 so the maximum value is 122/3 at x = 3 and the minimum value is 0 at x = −1, 0.

us uf

10. f (x) =

11. f (x) = 1 − sec2 x, f (x) = 0 for x in (−π/4, π/4) when x = 0; f (−π/4) = 1 − π/4, f (0) = 0, f (π/4) = π/4 − 1 so the maximum value is 1 − π/4 at x = −π/4 and the minimum value is π/4 − 1 at x = π/4.

13. f (x) = 1 + |9 − x2 | =

10 − x2 , |x| ≤ 3 , f (x) = −8 + x2 , |x| > 3

−2x, |x| < 3 2x, |x| > 3

x = 0, f (x) does not exist for x in (−5, 1) when x = −3 because

2, f (π) = 1

thus f (x) = 0 when

lim f (x) =

x→−3−

lim f (x) (see

x→−3+

Section 3.3); f (−5) = 17, f (−3) = 1, f (0) = 10, f (1) = 9 so the and the minimum value is 1 at x = −3. x ≤ 3/2 −4, x < 3/2 , f (x) = , f (x) does not exist when x > 3/2 4, x > 3/2

Theorem preceding Exercise 75, maximum value is 17 at x = −5 6 − 4x, 14. f (x) = |6 − 4x| = −6 + 4x,

√

12. f (x) = cos x + sin x, f (x)√= 0 for x in (0, π) when x = 3π/4; f (0) = −1, f (3π/4) = so the maximum value is 2 at x = 3π/4 and the minimum value is −1 at x = 0.

x = 3/2 thus 3/2 is the only critical point in (−3, 3); f (−3) = 18, f (3/2) = 0, f (3) = 6 so the maximum value is 18 at x = −3 and the minimum value is 0 at x = 3/2.

15. f (x) = 2x − 3; critical point x = 3/2. Minimum value f (3/2) = −13/4, no maximum. 16. f (x) = −4(x + 1); critical point x = −1. Maximum value f (−1) = 5, no minimum.

17. f (x) = 12x2 (1 − x); critical points x = 0, 1. Maximum value f (1) = 1, no minimum because lim f (x) = −∞. x→+∞

18. f (x) = 4(x3 + 1); critical point x = −1. Minimum value f (−1) = −3, no maximum. 19. No maximum or minimum because lim f (x) = +∞ and lim f (x) = −∞. x→+∞

x→−∞

20. No maximum or minimum because lim f (x) = +∞ and lim f (x) = −∞. x→−∞

x→+∞

21. f (x) = x(x + 2)/(x + 1)2 ; critical point x = −2 in (−5, −1). Maximum value f (−2) = −4, no minimum.

x→3+

x→3−

22. f (x) = −6/(x − 3)2 ; no critical points in [−5, 5] (x = 3 is not in the domain of f ). No maximum or minimum because lim f (x) = +∞ and lim f (x) = −∞. 10

x→+∞

-2

2 0

uh a

23. (x2 − 1)2 can never be less than zero because it is the square of x2 − 1; the minimum value is 0 for x = ±1, no maximum because lim f (x) = +∞.

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Chapter 5

24. (x−1)2 (x+2)2 can never be less than zero because it is the product of two squares; the minimum value is 0 for x = 1 or −2, no maximum because lim f (x) = +∞.

us uf

x→+∞

-3

5(8 − x) , f (x) = 0 when x = 8 and f (x) 3x1/3 does not exist when x = 0; f (−1) = 21, f (0) = 0, f (8) = 48, f (20) = 0 so the maximum value is 48 at x = 8 and the minimum value is 0 at x = 0, 20.

25. f (x) =

-1

26. f (x) = (2 − x2 )/(x2 + 2)√2 , f (x) = 0 for x in the interval (−1, √ 4) when x = 2; f (−1) = −1/3, √ f√( 2) = 2/4, √ f (4) = 2/9 so the maximum value is 2/4 at x = 2 and the minimum value is −1/3 at x = −1.

-0.4

27. f (x) = −1/x2 ; no maximum or minimum because there are no critical points in (0, +∞).

0.4

0.5

20 0

uh a

28. f (x) = (1 − x2 )/(x2 + 1)2 ; critical point x = 1. Maximum value f (1) = 1/2, minimum value 0 because f (x) is never less than zero on [0, +∞) and f (0) = 0.

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Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Exercise Set 5.5

29. f (x) = 2 sec x tan x − sec2 x = (2 sin x − 1)/ cos2 x, f (x) = 0√for x in (0, π/4) √ when x = π/6; f (0) = 2, value f (π/6) = 3, f (π/4) = 2 2−1 so the maximum √ is 2 at x = 0 and the minimum value is 3 at x = π/6.

30. f (x) = 2 sin x cos x − sin x = sin x(2 cos x − 1), f (x) = 0 for x in (−π, π) when x = 0, ±π/3; f (−π) = −1, f (−π/3) = 5/4, f (0) = 1, f (π/3) = 5/4, f (π) = −1 so the maximum value is 5/4 at x = ±π/3 and the minimum value is −1 at x = ±π.

1.5

31. f (x) = x2 (2x − 3)e−2x , f (x) = 0 for x in [1, 4] when x = 3/2; 27 if x = 1, 3/2, 4, then f (x) = e−2 , e−3 , 64e−8 ; 8 27 −3 e at x = 3/2, critical point at x = 3/2; absolute maximum of 8 −8 absolute minimum of 64e at x = 4

us uf

0 1.5

32. f (x) = (1 − ln 2x)/x2 , f (x) = 0 on [1, e] for x = e/2;

187

-1.5

0.2

4 0

0.76

if x = 1, e/2, e then f (x) = ln 2, 2/e, (ln 2 + 1)/e; 1 + ln 2 at x = e, absolute minimum of e absolute maximum of 2/e at x = e/2

1 0.64

33. f (x) = 1/2 + 2x/(x2 + 1),

√

f (x) = 0 on [−4, 0] for x = −2 ± 3 √ √ if x = −2 − 3, −2 + 3 then √ √ f (x) = −1 − 3/2 + ln 4 + ln(2 + 3) ≈ 0.84, √ √ −1 + 3/2 + ln 4 + ln(2 − 3) ≈ −0.06, √ absolute maximum at x = −2 − 3, √ absolute minimum at x = −2 + 3

-4

uh a

2.7

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-0.5

Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Chapter 5

34. f (x) = (x2 + 2x − 1)ex , √ f (x) = 0 at x = −2 + 2 √ and x = −1 − 2 (discard), √ √ √ f (−1 + 2) = (2 − 2 2)e−1+ 2) ≈ −1.25,

us uf

188

-2

absolute maximum at x = 2, f (2) = 3e2 ≈ 22.17, √ absolute minimum at x = −1 + 2 35. f (x) = −[cos(cos x)] sin x; f (x) = 0 if sin x = 0 or if cos(cos x) = 0. If sin x = 0, then x = π is the critical point in (0, 2π); cos(cos x) = 0 has no solutions because −1 ≤ cos x ≤ 1. Thus f (0) = sin(1), f (π) = sin(−1) = − sin(1), and f (2π) = sin(1) so the maximum value is sin(1) ≈ 0.84147 and the minimum value is − sin(1) ≈ −0.84147.

-20

4, x < 1 2x − 5, x > 1

1.5

0 0

37. f (x) =

36. f (x) = −[sin(sin x)] cos x; f (x) = 0 if cos x = 0 or if sin(sin x) = 0. If cos x = 0, then x = π/2 is the critical point in (0, π); sin(sin x) = 0 if sin x = 0, which gives no critical points in (0, π). Thus f (0) = 1, f (π/2) = cos(1), and f (π) = 1 so the maximum value is 1 and the minimum value is cos(1) ≈ 0.54030.

-1

so f (x) = 0 when x = 5/2, and f (x) does not exist when x = 1

x→1−

x→1+

because lim f (x) = lim f (x) (see Theorem preceding Exercise 75, Section 3.3); f (1/2) = 0, f (1) = 2, f (5/2) = −1/4, f (7/2) = 3/4 so the maximum value is 2 and the minimum value is −1/4.

38. f (x) = 2x + p which exists throughout the interval (0, 2) for all values of p so f (1) = 0 because f (1) is an extreme value, thus 2 + p = 0, p = −2. f (1) = 3 so 12 + (−2)(1) + q = 3, q = 4 thus f (x) = x2 − 2x + 4 and f (0) = 4, f (2) = 4 so f (1) is the minimum value.

uh a

39. sin 2x has a period of π, and sin 4x a period of π/2 so f (x) is periodic with period π. Consider the interval [0, π]. f (x) = 4 cos 2x + 4 cos 4x, f (x) = 0 when cos 2x + cos 4x = 0, but cos 4x = 2 cos2 2x − 1 (trig identity) so 2 cos2 2x + cos 2x − 1 = 0 (2 cos 2x − 1)(cos 2x + 1) = 0 cos 2x = 1/2 or cos 2x = −1.

From cos 2x = 1/2, 2x√= π/3 or 5π/3 so x = π/6 or 5π/6. √ From cos 2x = −1, 2x = π so x = π/2. f√ (0) = 0, f (π/6) = 3 3/2, f (π/2) = 0, f (5π/6) = −3 √ 3/2, f (π) = 0. The maximum value is 3 3/2 at x = π/6 + nπ and the minimum value is −3 3/2 at x = 5π/6 + nπ, n = 0, ±1, ±2, · · ·.

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Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Exercise Set 5.5

189

x x has a period of 6π, and cos a period of 4π, so f (x) has a period of 12π. Consider the 3 2 x x x x interval [0, 12π]. f (x) = − sin − sin , f (x) = 0 when sin + sin = 0 thus, by use of 3 2 3 2

a−b 5x x 5x a+b cos , 2 sin cos − = 0 so sin = 0 or the trig identity sin a + sin b = 2 sin 2 2 12 12 12

us uf

40. cos

x x 5x =0 = 0 to get x = 12π/5, 24π/5, 36π/5, 48π/5 and then solve cos = 0. Solve sin 12 12 12 to get x = 6π. The corresponding values of f (x) are −4.0450, 1.5450, 1.5450, −4.0450, 1, 5, 5 so the maximum value is 5 and the minimum value is −4.0450 (approximately).

cos

41. Let f (x) = x − sin x, then f (x) = 1 − cos x and so f (x) = 0 when cos x = 1 which has no solution for 0 < x < 2π thus the minimum value of f must occur at 0 or 2π. f (0) = 0, f (2π) = 2π so 0 is the minimum value on [0, 2π] thus x − sin x ≥ 0, sin x ≤ x for all x in [0, 2π].

42. Let h(x) = cos x − 1 + x2 /2. Then h(0) = 0, and it is suﬃcient to show that h (x) ≥ 0 for 0 < x < 2π. But h (x) = − sin x + x ≥ 0 by Exercise 41.

43. Let m = slope at x, then m = f (x) = 3x2 − 6x + 5, dm/dx = 6x − 6; critical point for m is x = 1, minimum value of m is f (1) = 2 −64 cos3 x + 27 sin3 x 64 cos x 27 sin x = + , f (x) = 0 when cos2 x sin2 x sin2 x cos2 x

44. (a) f (x) = −

27 cos3 x(tan3 x − 64/27) 27 cos x(tan3 x − 64/27) = ; sin2 x cos2 x sin2 x

f (x) =

27 sin3 x = 64 cos3 x, tan3 x = 64/27, tan x = 4/3 so the critical point is x = x0 where tan x0 = 4/3 and 0 < x0 < π/2. To test x0 ﬁrst rewrite f (x) as

if x < x0 then tan x < 4/3 and f (x) < 0, if x > x0 then tan x > 4/3 and f (x) > 0 so f (x0 ) is the minimum value. f has no maximum because lim f (x) = +∞. x→0+

(b) If tan x0 = 4/3 then (see ﬁgure)

sin x0 = 4/5 and cos x0 = 3/5 so f (x0 ) = 64/ sin x0 + 27/ cos x0 = 64/(4/5) + 27/(3/5) 5

x0 3

= 80 + 45 = 125

√ 2x(x3 − 24x2 + 192x − 640) 3 3 2 ; real root of x − 24x + 192x − 640 at x = 4(2 + 2). Since (x − 8)3

uh a

45. f (x) =

lim f (x) = lim f (x) = +∞ and there is only one relative extremum, it must be a minimum. x→+∞

x→8+

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190

Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Chapter 5

46. (a)

K −at ln(a/b) dC dC = = 0 at t = . This is the only stationary point and − be−bt so ae dt a−b dt a−b C(0) = 0, lim C(t) = 0, C(t) > 0 for 0 < t < +∞, so it is an absolute maximum.

us uf

0.7

(b)

t→+∞

10 0

47. The slope of the line is −1, and the slope of the tangent to y = −x2 is −2x so −2x = −1, x = 1/2. The line lies above the curve so the vertical distance is given by F (x) = 2 − x + x2 ; F (−1) = 4, F (1/2) = 7/4, F (3/2) = 11/4. The point (1/2, −1/4) is closest, the point (−1, −1) farthest.

48. The slope of the line is 4/3; and the slope of the tangent to y = x3 is 3x2 so 3x2 = 4/3, x2 = 4/9, x = ±2/3. The line lies below the curve so the vertical distance is given by F (x) = x3 − 4x/3 + 1; F (−1) = 4/3, F (−2/3) = 43/27, F (2/3) = 11/27, F (1) = 2/3. The closest point is (2/3, 8/27), the farthest is (−2/3, −8/27).

49. The absolute extrema of y(t) can occur at the endpoints t = 0, 12 or when dy/dt = 2 sin t = 0, i.e. t = 0, 12, kπ, k = 1, 2, 3; the absolute maximum is y = 4 at t = π, 3π; the absolute minimum is y = 0 at t = 0, 2π.

50. (a) The absolute extrema of y(t) can occur at the endpoints t = 0, 2π or when dy/dt = 2 cos 2t − 4 sin t cos t = 2 cos 2t − 2 sin 2t = 0, t = 0, 2π, π/8, 5π/8, 9π/8, 13π/8; the absolute maximum is y = 3.4142 at t = π/8, 9π/8; the absolute minimum is y = 0.5859 at t = 5π/8, 13π/8. (b) The absolute extrema of x(t) occur at the endpoints t = 0, 2π or when dx 2 sin t + 1 = 0, t = 7π/6, 11π/6. The absolute maximum is x = 0.5774 at t = 11π/6 =− dt (2 + sin t)2 and the absolute minimum is x = −0.5774 at t = 7π/6.

51. f (x) = 2ax + b; critical point is x = −

b 2a

b is the minimum value of f , but f (x) = 2a > 0 so f − 2a 2 b b −b2 + 4ac b f − =a − +c= thus f (x) ≥ 0 if and only if +b − 2a 2a 2a 4a b −b2 + 4ac f − ≥ 0, ≥ 0, −b2 + 4ac ≥ 0, b2 − 4ac ≤ 0 2a 4a

uh a

52. Use the proof given in the text, replacing “maximum” by “minimum” and “largest” by “smallest” and reversing the order of all inequality symbols.

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Exercise Set 5.6

191

EXERCISE SET 5.6

us uf

1. Let x = one number, y = the other number, and P = xy where x + y = 10. Thus y = 10 − x so P = x(10 − x) = 10x − x2 for x in [0, 10]. dP/dx = 10 − 2x, dP/dx = 0 when x = 5. If x = 0, 5, 10 then P = 0, 25, 0 so P is maximum when x = 5 and, from y = 10 − x, when y = 5.

(a) z is as large as possible when one number is 0 and the other is 1. (b) z is as small as possible when both numbers are 1/2.

2. Let x and y be nonnegative numbers and z the sum of their squares, then z = x2 + y 2 . But x + y = 1, y = 1 − x so z = x2 + (1 − x)2 = 2x2 − 2x + 1 for 0 ≤ x ≤ 1. dz/dx = 4x − 2, dz/dx = 0 when x = 1/2. If x = 0, 1/2, 1 then z = 1, 1/2, 1 so

Stream

4. A = xy where x + 2y = 1000 so y = 500 − x/2 and A = 500x − x2 /2 for x in [0, 1000]; dA/dx = 500 − x, dA/dx = 0 when x = 500. If x = 0 or 1000 then A = 0, if x = 500 then A = 125, 000 so the area is maximum when x = 500 ft and y = 500 − 500/2 = 250 ft.

3. If y = x + 1/x for 1/2 ≤ x ≤ 3/2 then dy/dx = 1 − 1/x2 = (x2 − 1)/x2 , dy/dx = 0 when x = 1. If x = 1/2, 1, 3/2 then y = 5/2, 2, 13/6 so (a) y is as small as possible when x = 1. (b) y is as large as possible when x = 1/2.

x Heavy-duty

Standard

5. Let x and y be the dimensions shown in the ﬁgure and A the area, then A = xy subject to the cost condition 3(2x) + 2(2y) = 6000, or y = 1500 − 3x/2. Thus A = x(1500 − 3x/2) = 1500x − 3x2 /2 for x in [0, 1000]. dA/dx = 1500−3x, dA/dx = 0 when x = 500. If x = 0 or 1000 then A = 0, if x = 500 then A = 375, 000 so the area is greatest when x = 500 ft and (from y = 1500 − 3x/2) when y = 750 ft.

y 10

uh a

7. Let x, y, and z be as shown in the ﬁgure and A the area of the rectangle, then A = xy and, by similar triangles, z/10 = y/6, z = 5y/3; also x/10 = (8 − z)/8 = (8 − 5y/3)/8 thus y = 24/5 − 12x/25 so A = x(24/5 − 12x/25) = 24x/5 − 12x2 /25 for x in [0, 10]. dA/dx = 24/5 − 24x/25, dA/dx = 0 when x = 5. If x = 0, 5, 10 then A = 0, 12, 0 so the area is greatest when x = 5 in. and y = 12/5 in.

x y

6. Let x and y be the dimensions shown in the ﬁgure and A the area of the rectangle, then A = xy and, by similar triangles, x/6 = (8 − y)/8, y = 8 − 4x/3 so A = x(8 − 4x/3) = 8x − 4x2 /3 for x in [0, 6]. dA/dx = 8 − 8x/3, dA/dx = 0 when x = 3. If x = 0, 3, 6 then A = 0, 12, 0 so the area is greatest when x = 3 in and (from y = 8 − 4x/3) y = 4 in.

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z 8

x 6

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Chapter 5

3 8. A = (2x)y = 2xy where y = 16 − x2 so A = 32x − 2x √ for 2 0 ≤ x ≤ 4; √ = 0 when x = 4/ 3. If √ dA/dx = 32 − 6x , dA/dx x = 0, 4/ 3,√4 then A = 0, 256/(3 3), 0 so the area is largest when x = 4/ 3 and y =√32/3. The dimensions of the rectangle with largest area are 8/ 3 by 32/3.

us uf

√ 9. A = xy√where x2 + y 2 = 202 = 400 so y = 400 − x2 and A = x 400 − x2 for 0 √ ≤ x ≤ 20; 2 2 , dA/dx = 0 when 400 − x√ dA/dx = 2(200 − x )/ √ √ the x = 200 = 10 2. If x = 0, 10√ 2, 20 then√A = 0, 200, 0 so √ area is maximum when x = 10 2 and y = 400 − 200 = 10 2.

-4

x R

x 2

10. Let x and y be the dimensions shown in the ﬁgure, then the area of the rectangle is A = xy. x 2 But + y 2 = R2 , thus 2 1 2 y = R2 − x2 /4 = 4R − x2 so 2 1 A = x 4R2 − x2 for 0 ≤ x ≤ 2R. 2 √ dA/dx = (2R2 − x2 )/ 4R2 − x2 , dA/dx = 0 when √ √ 2 x = 2R. If x = 0, 2R, 2R then √ A = 0, R , 0√so the greatest area occurs when x = 2R and y = 2R/2.

11. Let x = length of each side that uses the $1 per foot fencing, y = length of each side that uses the $2 per foot fencing. The cost is C = (1)(2x) + (2)(2y) = 2x + 4y, but A = xy = 3200 thus y = 3200/x so C = 2x + 12800/x for x > 0, dC/dx = 2 − 12800/x2 , dC/dx = 0 when x = 80, d2 C/dx2 > 0 so

C is least when x = 80, y = 40.

y x

12. A = xy where 2x + 2y = p so y = p/2 − x and A = px/2 − x2 for x in [0, p/2]; dA/dx = p/2 − 2x, dA/dx = 0 when x = p/4. If x = 0 or p/2 then A = 0, if x = p/4 then A = p2 /16 so the area is maximum when x = p/4 and y = p/2 − p/4 = p/4, which is a square.

uh a

13. Let x and y be the dimensions of a rectangle; the perimeter is p = 2x + 2y. But A = xy thus y =√ A/x so p = 2x + 2A/x for x > 0, dp/dx = 2 − 2A/x2 = 2(x2 −√A)/x2 , dp/dx √ = 0 when x = A, d2 p/dx2 = 4A/x3 > 0 if x > 0 so p is a minimum when x = A and y = A and thus the rectangle is a square.

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Exercise Set 5.6

us uf

cut

(12 − x) π+4 2 3 x + = x − x + 9 for 0 ≤ x ≤ 12. A= 4π 16 16π 2 12π 12π π+4 dA 3 dA = 0 when x = . If x = 0, , 12 = x− , π+4 π+4 dx 8π 2 dx 36 36 then A = 9, , so the sum of the enclosed areas is π+4 π

14. With x, y, r, and s as shown in the ﬁgure, the sum of x and the enclosed areas is A = πr2 + s2 where r = 2π y s = because x is the circumference of the circle and 4 y2 x2 + y is the perimeter of the square, thus A = . 4π 16 But x + y = 12, so y = 12 − x and

193

(a) a maximum when x = 12 in. (when all of the wire is used for the circle) (b) a minimum when x = 12π/(π + 4) in.

dN = 250(20 − t)e−t/20 = 0 at t = 20, N (0) = 125,000, N (20) ≈ 161,788, and dt N (100) ≈ 128,369; the absolute maximum is N = 161788 at t = 20, the absolute minimum is N = 125,000 at t = 0. d2 N dN occurs when (b) The absolute minimum of = 12.5(t − 40)e−t/20 = 0, t = 40. dt dt2

15. (a)

1 [p − (2 + π)r] so 2 A = r[p − (2 + π)r] + πr2 /2

16. The area of the window is A = 2rh + πr2 /2, the perimeter is p = 2r + 2h + πr thus

= pr − (2 + π/2)r2 for 0 ≤ r ≤ p/(2 + π),

dA/dr = p − (4 + π)r, dA/dr = 0 when

r = p/(4 + π) and d2 A/dr2 < 0, so A is maximum when r = p/(4 + π).

12 x x

x x

12 - 2x

12 x x

x x 12 - 2x

17. V = x(12 − 2x)2 for 0 ≤ x ≤ 6; dV /dx = 12(x − 2)(x − 6), dV /dx = 0 when x = 2 for 0 < x < 6. If x = 0, 2, 6 then V = 0, 128, 0 so the volume is largest when x = 2 in.

uh a

18. The dimensions of the box will be (k − 2x) by (k − 2x) by x so V = (k − 2x)2 x = 4x3 − 4kx2 + k 2 x for x in [0, k/2]. dV /dx = 12x2 − 8kx + k 2 = (6x − k)(2x − k), dV /dx = 0 for x in (0, k/2) when x = k/6. If x = 0, k/6, k/2 then V = 0, 2k 3 /27, 0 so V is maximum when x = k/6. The squares should have dimensions k/6 by k/6.

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Chapter 5

us uf

19. Let x be the length of each side of a square, then V = x(3 − 2x)(8 − 2x) = 4x3 − 22x2 + 24x for 0 ≤ x ≤ 3/2; dV /dx = 12x2 − 44x + 24 = 4(3x − 2)(x − 3), dV /dx = 0 when x = 2/3 for 0 < x < 3/2. If x = 0, 2/3, 3/2 then V = 0, 200/27, 0 so the maximum volume is 200/27 ft3 . 20. Let x = length of each edge of base, y = height. The cost is C = (cost of top and bottom) + (cost of sides) = (2)(2x2 ) + (3)(4xy) = 4x2 + 12xy, but V = x2 y = 2250 thus y√= 2250/x2 so C = 4x2 + 27000/x for x > 0, dC/dx = 8x − 27000/x2 , dC/dx = 0 when x = 3 3375 = 15, d2 C/dx2 > 0 so C is least when x = 15, y = 10.

22. Let x and y be the dimensions shown in the ﬁgure and V the volume, then V = x2 y. The amount of material is to be 1000 ft2 , thus (area of base) + (area 1000 − x2 of sides) = 1000, x2 + 4xy = 1000, y = so 4x 2 √ 1 1000 − x = (1000x−x3 ) for 0 < x ≤ 10 10. V = x2 4x 4

21. Let x = length of each edge of base, y = height, k = $/cm2 for the sides. The cost is C = (2k)(2x2 ) + (k)(4xy) = 4k(x2 + xy), but V = x2 y = 2000 thus y = 2000/x2 so C = 4k(x2 + 2000/x) for x > 0 dC/dx = 4k(2x − 2000/x2 ), dC/dx = 0 when √ x = 3 1000 = 10, d2 C/dx2 > 0 so C is least when x = 10, y = 20.

x x

1 dV dV =0 = (1000 − 3x2 ), 4 dx dx when x = 1000/3 = 10 10/3. √ 5000 10/3, 0; If x = 0, 10 10/3, 10 10 then V = 0, 3 the volume is greatest for x = 10 10/3 ft and y = 5 10/3 ft.

V , so 23. Let x = height and width, y = length. The surface area is S = 2x2 + 3xy where x2 y = 3 2 dS/dx = 4x − 3V /x , dS/dx = 0 when x = 3V /4, y = V /x2 and S = 2x2 + 3V /x for x > 0; 4 3 3V 3 3V d2 S/dx2 > 0 so S is minimum when x = ,y= . 4 3 4

uh a

24. Let r and h be the dimensions shown in the ﬁgure, then the volume of the inscribed cylinder is V = πr2 h. But 2 h h2 2 r + = R2 thus r2 = R2 − 2 4 2 h h3 2 2 h=π R h− so V = π R − 4 4 dV dV 3 for 0 ≤ h ≤ 2R. =0 = π R 2 − h2 , dh 4 dh √ √ when h = 2R/ 3. If h = 0, 2R/ 3, 2R

h h 2

4π then V = 0, √ R3 , 0 so the volume is largest 3 3 √ when h = 2R/ 3 and r = 2/3R.

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Exercise Set 5.6

us uf

25. Let r and h be the dimensions shown in the ﬁgure, then the surface area is S = 2πrh + 2πr2 . 2 √ h 2 But r + = R2 thus h = 2 R2 − r2 so 2 √ S = 4πr R2 − r2 + 2πr2 for 0 ≤ r ≤ R,

dS 4π(R2 − 2r2 ) dS = √ + 4πr; = 0 when 2 2 dr dr R −r R2 − 2r2 √ = −r R2 − r 2 √ R2 − 2r2 = −r R2 − r2

195

h 2

(1)

R4 − 4R2 r2 + 4r4 = r2 (R2 − r2 ) √

5r4 − 5R2 r2 + R4 = 0

√ − 5± 5 5R ± 5± 5 2 and using the quadratic formula r2 = = R , r = R, of 10 10 10 √ √ √ 5+ 5 5+ 5 which only r = R satisﬁes (1). If r = 0, R, 0 then S = 0, (5 + 5)πR2 , 2πR2 so 10 10 √ √ 5+ 5 5− 5 2 2 the surface area is greatest when r = R and, from h = 2 R − r , h = 2 R. 10 10 25R4

20R4

√

r H h

26. Let R and H be the radius and height of the cone, and r and h the radius and height of the cylinder (see ﬁgure), then the volume of the cylinder is V = πr2 h. r H −h = thus By similar triangles (see ﬁgure) H R H H H h = (R − r) so V = π (R − r)r2 = π (Rr2 − r3 ) R R R H dV H = π (2Rr−3r2 ) = π r(2R−3r), for 0 ≤ r ≤ R. dr R R dV = 0 for 0 < r < R when r = 2R/3. If dr r = 0, 2R/3, R then V = 0, 4πR2 H/27, 0 so the maxi-

4 1 2 4πR2 H 4 = πR H = · (volume of mum volume is 27 9 3 9 cone).

2 27. From (13), S = 2πr2 + 2πrh. But V = πr h thus h = V /(πr2 ) and so S = 2πr2 + 2V /r for r > 0. 3 2 r = V /(2π). Since d2 S/dr2 = 4π + 4V /r3 > 0, the minimum dS/dr = 4πr − 2V /r , dS/dr = 0 if 3 surface area is achieved when r = V /2π and so h = V /(πr2 ) = [V /(πr3 )]r = 2r.

1 S − 2πr2 , V = (Sr − 2πr3 ) for r > 0. 2πr 2 1 d2 V dV = (S − 6πr2 ) = 0 if r = S/(6π), = −6πr < 0 so V is maximum when dr 2 dr2 S − 2πr2 S − 2πr2 S − S/3 = r = 2r, thus the height is equal to the r= r = S/(6π) and h = 2πr 2πr2 S/3 diameter of the base.

uh a

28. V = πr2 h where S = 2πr2 + 2πrh so h =

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Chapter 5

29. The surface area is S = πr2 + 2πrh where V = πr2 h = 500 so h = 500/(πr2 ) and S = πr2 + 1000/r for r > 0; dS/dr = 2πr − 1000/r2 = (2πr3 − 1000)/r2 , dS/dr = 0 when r = 3 500/π, d2 S/dr2 > 0 for r > 0 so S is minimum when r = 3 500/π cm and 500 π 2/3 500 h= 2 = πr π 500 = 3 500/π cm

us uf

30. The total area of material used is A = Atop + Abottom + Aside = (2r)2 + (2r)2 + 2πrh = 8r2 + 2πrh. The volume is V = πr2 h thus h = V /(πr2 ) so A = 8r2 + 2V /r √ for r > 0, dA/dr = 16r − 2V /r2 = 2(8r3 − V )/r2 , dA/dr = 0 when r = 3 V /2. This is the only critical point, √ r r π d2 A/dr2 > 0 there so the least material is used when r = 3 V /2, = = r3 and, for 2 h V /(πr ) V √ r πV π 3 r = V /2, = = . h V 8 8

31. Let x be the length of each side of the squares and y the height of the frame, then the volume is V = x2 y. The total length of the wire is L thus 8x + 4y = L, y = (L − 8x)/4 so V = x2 (L − 8x)/4 = (Lx2 − 8x3 )/4 for 0 ≤ x ≤ L/8. dV /dx = (2Lx − 24x2 )/4, dV /dx = 0 for 0 < x < L/8 when x = L/12. If x = 0, L/12, L/8 then V = 0, L3 /1728, 0 so the volume is greatest when x = L/12 and y = L/12.

32. (a) Let x = diameter of the sphere, y = length of an edge of the cube. The combined volume is 1 (S − πx2 )1/2 V = πx3 + y 3 and the surface area is S = πx2 + 6y 2 = constant. Thus y = 6 61/2 S π 3 (S − πx2 )3/2 and V = x + for 0 ≤ x ≤ ; 6 π 63/2 √ 3π π dV dV π = 0 when x = 0, or when = x2 − 3/2 x(S − πx2 )1/2 = √ x( 6x − S − πx2 ). dx dx 2 6 2 6 √ √ S S S S 2 2 2 2 , x = . If x = 0, , , 6x = S − πx , 6x = S − πx , x = 6+π 6+π 6+π π S 3/2 S S 3/2 S 3/2 then V = 3/2 , √ , and hence when , √ so that V is smallest when x = 6 + π π 6 6 6 6 + π S y= , thus x = y. 6+π

(b) From Part (a), the sum of the volumes is greatest when there is no cube.

uh a

33. Let h and r be the dimensions shown in the ﬁgure, 1 then the volume is V = πr2 h. But r2 + h2 = L2 thus 3 1 1 r2 = L2 − h2 so V = π(L2 − h2 )h = π(L2 h − h3 ) 3 3 1 dV dV = π(L2 − 3h2 ). for 0 ≤ h ≤ L. = 0 when dh 3 dh √ √ 2π h = L/ 3. If h = 0, L/ 3, 0 then V = 0, √ L3 , 0 so 9 3√ the volume is as large as possible when h = L/ 3 and r = 2/3L.

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Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

us uf

34. Let r and h be the radius and height of the cone (see ﬁgure). The slant height of any such cone will be R, the radius of the circular sheet. Refer to the solution of 2π Exercise 33 to ﬁnd that the largest volume is √ R3 . 9 3

197

Exercise Set 5.6

√

To simplify the computations let S = A2 , 900 900 S = π 2 r2 r2 + 2 4 = π 2 r4 + 2 for r > 0, π r r

35. The area of the paper is A = πrL = πr r2 + h2 , but 1 V = πr2 h = 10 thus h = 30/(πr2 ) 3 so A = πr r2 + 900/(π 2 r4 ).

dS 1800 4(π 2 r6 − 450) = 4π 2 r3 − 3 = , dS/dr = 0 when dr r r3 r = 6 450/π 2 , d2 S/dr2 > 0, so S and hence A is least 30 3 π 2 /450 cm. when r = 6 450/π 2 cm, h = π

1 hb. By similar trian2 R 2Rh b/2 =√ ,b= √ gles (see ﬁgure) h h2 − 2Rh h2 − 2Rh Rh2 dA Rh2 (h − 3R) , for h > 2R, = 2 dh (h − 2Rh)3/2 h2 − 2Rh

so A = √

36. The area of the triangle is A =

h−R

h R

dA = 0 for h > 2R when h = 3R, by the ﬁrst derivadh tive test √ A is minimum when h = 3R. If h = 3R then b = 2 3R (the triangle is equilateral).

37. The volume of the cone is V =

1 2 πr h. By similar tri3 h−R

h2 − 2Rh

h h 1 2 1 = πR2 for h > 2R, πR 2 h − 2Rh 3 h − 2R 3

so V =

b/2 b

R r Rh = √ ,r= √ 2 2 h h − 2Rh h − 2Rh

angles (see ﬁgure)

h2 − 2Rh

R R

uh a

1 h(h − 4R) dV dV = 0 for h > 2R when = πR2 , 3 (h − 2R)2 dh dh

h = 4R, by the ﬁrst derivative test √ V is minimum when h = 4R. If h = 4R then r = 2R.

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Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Chapter 5

38. The area is (see ﬁgure) 1 A = (2 sin θ)(4 + 4 cos θ) 2 = 4(sin θ + sin θ cos θ) for 0 ≤ θ ≤ π/2; dA/dθ = 4(cos θ − sin2 θ + cos2 θ) = 4(cos θ − [1 − cos2 θ] + cos2 θ) = 4(2 cos2 θ + cos θ − 1) = 4(2 cos θ − 1)(cos θ + 1) dA/dθ =√0 when θ = π/3 for 0 < θ < π/2. √ If θ = 0, π/3, π/2 then A = 0, 3 3, 4 so the maximum area is 3 3. 39. Let b and h be the dimensions shown in the ﬁgure, 1 then the cross-sectional area is A = h(5 + b). But 2 h = 5 sin θ and b = 5 + 2(5 cos θ) = 5 + 10 cos θ so 5 A = sin θ(10 + 10 cos θ) = 25 sin θ(1 + cos θ) for 2 0 ≤ θ ≤ π/2.

2 sin θ

us uf

2 cos θ

5 cos θ

h = 5 sin θ

dA/dθ = −25 sin2 θ + 25 cos θ(1 + cos θ) = 25(− sin2 θ + cos θ + cos2 θ) = 25(−1 + cos2 θ + cos θ + cos2 θ) = 25(2 cos2 θ + cos θ − 1) = 25(2 cos θ − 1)(cos θ + 1). dA/dθ = 0 for 0 < θ < π/2 when√cos θ = 1/2, θ = π/3. If θ = 0, π/3, π/2 then A = 0, 75 3/4, 25 so the crosssectional area is greatest when θ = π/3.

4 cos θ

198

cos φ , k the constant of proportionality. If h is the height of the lamp above the table then -2 √ h h dI dI r2 − 2h2 for h > 0, , cos φ = h/- and - = h2 + r2 so I = k 3 = k 2 = k =0 2 3/2 2 2 5/2 dh dh (h + r ) (h + r ) √ √ when h = r/ 2, by the ﬁrst derivative test I is maximum when h = r/ 2.

40. I = k

41. Let L, L1 , and L2 be as shown in the ﬁgure, then L = L1 + L2 = 8 csc θ + sec θ, L2 L L1 8 θ

dL = −8 csc θ cot θ + sec θ tan θ, 0 < θ < π/2 dθ 8 cos θ −8 cos3 θ + sin3 θ sin θ =− ; = + cos2 θ sin2 θ sin2 θ cos2 θ

x = number of steers per acre w = average market weight per steer T = total market weight per acre then T = xw where w = 2000 − 50(x − 20) = 3000 − 50x so T = x(3000 − 50x) = 3000x − 50x2 for 0 ≤ x ≤ 60, dT /dx = 3000 − 100x and dT /dx = 0 when x = 30. If x = 0, 30, 60 then T = 0, 45000, 0 so the total market weight per acre is largest when 30 steers per acre are allowed.

uh a

42. Let

dL = 0 if sin3 θ = 8 cos3 θ, tan3 θ = 8, tan θ = 2 which gives dθ the absolute minimum for L because lim+ L = lim − L = +∞. θ→0 θ→π/2 √ √ √ √ √ If tan θ = 2, then csc θ = 5/2 and sec θ = 5 so L = 8( 5/2) + 5 = 5 5 ft.

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Exercise Set 5.6

199

43. (a) The daily proﬁt is

P = (revenue) − (production cost) = 100x − (100, 000 + 50x + 0.0025x2 ) = −100, 000 + 50x − 0.0025x2

us uf

for 0 ≤ x ≤ 7000, so dP/dx = 50 − 0.005x and dP/dx = 0 when x = 10, 000. Because 10,000 is not in the interval [0, 7000], the maximum profit must occur at an endpoint. When x = 0, P = −100, 000; when x = 7000, P = 127, 500 so 7000 units should be manufactured and sold daily. (b) Yes, because dP/dx > 0 when x = 7000 so profit is increasing at this production level. (c) dP/dx = 15 when x = 7000, so P (7001) − P (7000) ≈ 15, and the marginal profit is $15.

44. (a) R(x) = px but p = 1000 − x so R(x) = (1000 − x)x (b) P (x) = R(x) − C(x) = (1000 − x)x − (3000 + 20x) = −3000 + 980x − x2 (c) P (x) = 980 − 2x, P (x) = 0 for 0 < x < 500 when x = 490; test the points 0, 490, 500 to ﬁnd that the proﬁt is a maximum when x = 490. (d) P (490) = 237,100 (e) p = 1000 − x = 1000 − 490 = 510.

45. The proﬁt is

P = (proﬁt on nondefective) − (loss on defective) = 100(x − y) − 20y = 100x − 120y

but y = 0.01x + 0.00003x2 so P = 100x − 120(0.01x + 0.00003x2 ) = 98.8x − 0.0036x2 for x > 0, dP/dx = 98.8 − 0.0072x, dP/dx = 0 when x = 98.8/0.0072 ≈ 13, 722, d2 P/dx2 < 0 so the proﬁt is maximum at a production level of about 13,722 pounds. 46. The total cost C is

=c·

C = c · (hours to travel 3000 mi at a speed of v mi/h)

3000 3000 = (a + bv n ) = 3000(av −1 + bv n−1 ) for v > 0, v v

dC/dv = 3000[−av −2 + b(n − 1)v n−2 ] = 3000[−a + b(n − 1)v n ]/v 2 , 1/n a dC/dv = 0 when v = . This is the only critical point and dC/dv changes sign from b(n − 1) 1/n a − to + at this point so the total cost is least when v = mi/h. b(n − 1)

uh a

√ 47. The distance between the particles is D = (1 − t − t)2 + (t − 2t)2 = 5t2 − 4t + 1 for t ≥ 0. For convenience, we minimize D2 instead, so D2 = 5t2 − 4t + 1, dD2 /dt = 10t − 4, which is 0 when t = 2/5.√d2 D2 /dt2 > 0 so D2 and hence D is minimum when t = 2/5. The minimum distance is D = 1/ 5. √ 48. The distance between the particles is D = (2t − t)2 + (2 − t2 )2 = t4 − 3t2 + 4 for t ≥ 0. For 3 2 − 3/2), which convenience we minimize D2 instead so D2 = t4 − 3t2 + 4, dD2 /dt = 4t − 6t = 4t(t 2 2 2 2 2 is 0 for t > 0 when t = 3/2. d D /dt = 12t√− 6 > 0 when t = 3/2 so D and hence D is minimum there. The minimum distance is D = 7/2.

49. Let P (x, y) be a point on the curve x2 + y 2 = 1. The distance between P (x, y) and P0 (2, 0) is √ D = (x − 2)2 + y 2 , but y 2 = 1 − x2 so D = (x − 2)2 + 1 − x2 = 5 − 4x for −1 ≤ x ≤ 1, dD 2 = −√ which has no critical points for −1 < x < 1. If x = −1, 1 then D = 3, 1 so the dx 5 − 4x closest point occurs when x = 1 and y = 0.

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200

Chapter 5

us uf

√ 50. Let P (x, y) be a point on y = x, then the √ distance D between P and (2, 0) is D = (x − 2)2 + y 2 = (x − 2)2 + x = x2 − 3x + 4, for 0 ≤ x ≤ 3. For convenience we ﬁnd the = x2 − 3x + 4, dD2 /dx = 2x − 3 =√0 when x = 3/2. If x = 0, 3/2, 3 extrema for D2 instead, so D2 √ 2 then D = 4, 7/4, 4 so D = 2, 7/2, 2. The points (0, 0) and (3, 3) are at the greatest distance, and (3/2, 3/2) the shortest distance from (2, 0). 51. Let (x, y) be a point on the curve, then the square of the distance between (x, y) and (0, 2) is S = x2 + (y − 2)2 where x2 − y 2 = 1, x2 = y 2 + 1 so S = (y 2 + 1) + (y − 2)2 = 2y 2 − 4y + 5 for any y,√dS/dy = 4y − 4, dS/dy = 0 when y = 1, d2 S/dy 2 > 0 so S is least when y = 1 and x = ± 2.

52. The square of the distance between a point (x, y) on the curve and the point (0, 9) is S = x2 + (y − 9)2 where x = 2y 2 so S = 4y 4 + (y − 9)2 for any y, dS/dy = 16y 3 + 2(y − 9) = 2(8y 3 + y − 9), dS/dy = 0 when y = 1 (which is the only real solution), d2 S/dy 2 > 0 so S is least when y = 1, x = 2.

53. If P (x0 , y0 ) is on the curve y = 1/x2 , then y0 = 1/x20 . At P the slope of the tangent line is −2/x30 1 2 2 3 so its equation is y − 2 = − 3 (x − x0 ), or y = − 3 x + 2 . The tangent line crosses the y-axis x0 x0 x0 x0

3 9 3 9 , and the x-axis at x0 . The length of the segment then is L = + x20 for x0 > 0. For 2 4 2 x0 4 x0 2 9 9 dL 36 9 9(x60 − 8) convenience, we minimize L2 instead, so L2 = 4 + x20 , = − 5 + x0 = , which x0 4 dx0 x0 2 2x50 √ d2 L2 √ is 0 when x60 = 8, x0 = 2. > 0 so L2 and hence L is minimum when x0 = 2, y0 = 1/2. 2 dx0

54. If P (x0 , y0 ) is on the curve y = 1 − x2 , then y0 = 1 − x20 . At P the slope of the tangent line is −2x0 so its equation is y − (1 − x20 ) = −2x0 (x − x0 ), or y = −2x0 x + x20 + 1. The y-intercept is 1 x20 + 1 and the x-intercept is (x0 + 1/x0 ) so the area A of the triangle is 2 1 2 1 3 A = (x0 + 1)(x0 + 1/x0 ) = (x0 + 2x0 + 1/x0 ) for 0 ≤ x0 ≤ 1. 4 4 1 1 dA/dx0 = (3x20 + 2 − 1/x20 ) = (3x40 + 2x20 − 1)/x20 which is 0 when x20 = −1 (reject), or when 4 4 √ √ 1 x20 = 1/3 so x0 = 1/ 3. d2 A/dx20 = (6x0 + 2/x30 ) > 0 at x0 = 1/ 3 so a relative minimum and 4 hence the absolute minimum occurs there. 55. At each point (x, y) on the curve the slope of the tangent line is m =

dy 2x for any =− dx (1 + x2 )2

√ dm 2(3x2 − 1) dm , = = 0 when x = ±1/ 3, by the ﬁrst derivative test the only relative dx (1 + x2 )3 dx √ maximum occurs at x = −1/ 3, which is the absolute maximum because lim m = 0. The x→±∞ √ tangent line has greatest slope at the point (−1/ 3, 3/4).

uh a

56. Let x be how far P is upstream from where the man starts (see ﬁgure), then the total time to reach T is

t = (time from M to P ) + (time from P to T ) √ x2 + 1 1 − x = + for 0 ≤ x ≤ 1, rR rW where rR and rW are the rates at which he can row and walk, respectively.

1 x P

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Exercise Set 5.6

201

√

x dt x2 + 1 1 − x dt 1 + , = √ = 0 when 5x = 3 x2 + 1, − so 3 5 dx dx 3 x2 + 1 5 √ 25x2 = 9(x2 + 1), x2 = 9/16, x = 3/4. If x = 0, 3/4, 1 then t = 8/15, 7/15, 2/3 so the time is a minimum when x = 3/4 mile. √ x2 + 1 1 1 − x dt x dt (b) t = − so + , = √ = 0 when x = 4/3 which is not in the 2 4 5 dx 5 dx 4 x +1 interval [0, 1]. Check the endpoints to ﬁnd that the time is a minimum when x = 1 (he should row directly to the town).

us uf

(a) t =

x y 8x =√ so ,y=√ 2 2 8 x − 16 x − 16 128 dL 8x =1− 2 , for x > 4, L=x+ √ dx (x − 16)3/2 x2 − 16 dL = 0 when dx

57. With x and y as shown in the ﬁgure, the maximum length of pipe will be the smallest value of L = x + y. By similar triangles

x2 - 16

(x2 − 16)3/2 = 128 x2 − 16 = 1282/3 = 16(22/3 ) x2 = 16(1 + 22/3 ) x = 4(1 + 22/3 )1/2 ,

d2 L/dx2 = 384x/(x2 − 16)5/2 > 0 if x > 4 so L is smallest when x = 4(1 + 22/3 )1/2 . For this value of x, L = 4(1 + 22/3 )3/2 ft.

s = (x1 − x ¯)2 + (x2 − x ¯)2 + · · · + (xn − x ¯)2 , ¯) − 2(x2 − x ¯) − · · · − 2(xn − x ¯), ds/d¯ x = −2(x1 − x ds/d¯ x = 0 when

58.

¯) + (x2 − x ¯) + · · · + (xn − x ¯) = 0 (x1 − x (x1 + x2 + · · · xn ) − (¯ x+x ¯ + ··· + x ¯) = 0 x= 0 (x1 + x2 + · · · + xn ) − n¯ 1 x ¯ = (x1 + x2 + · · · + xn ), n x2 = 2 + 2 + · · · + 2 = 2n > 0, so s is minimum when x ¯= d2 s/d¯

1 (x1 + x2 + · · · + xn ). n

kS 8kS + if 0 < x < 90; x2 (90 − x)2

59. Let x = distance from the weaker light source, I = the intensity at that point, and k the constant of proportionality. Then

uh a

16kS 2kS[8x3 − (90 − x)3 ] kS(x − 30)(x2 + 2700) dI 2kS = = 18 , =− 3 + dx x (90 − x)3 x3 (90 − x)3 x3 (x − 90)3 dI dI > 0 if x > 30, so the intensity is minimum at a which is 0 when x = 30; < 0 if x < 30, and dx dx distance of 30 cm from the weaker source.

60. If f (x0 ) is a maximum then f (x) ≤ f (x0 ) for all x in some open interval containing x0 thus √ f (x) ≤ f (x0 ) because x is an increasing function, so f (x0 ) is a maximum of f (x) at x0 . The proof is similar for a minimum value, simply replace ≤ by ≥.

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Chapter 5

61. Î¸ = Ď&#x20AC; â&#x2C6;&#x2019; (Îą + Î˛)

5â&#x2C6;&#x2019;x = Ď&#x20AC; â&#x2C6;&#x2019; cot (x â&#x2C6;&#x2019; 2) â&#x2C6;&#x2019; cot , 4 1 dÎ¸ â&#x2C6;&#x2019;1/4 = + dx 1 + (x â&#x2C6;&#x2019; 2)2 1 + (5 â&#x2C6;&#x2019; x)2 /16 â&#x2C6;&#x2019;1

B(5, 4)

â&#x2C6;&#x2019;1

P(x, 0) A(2, 1) â?Ł

â?Ş

us uf

202

â?¤

5 3(x â&#x2C6;&#x2019; 2x â&#x2C6;&#x2019; 7) x-2 5-x =â&#x2C6;&#x2019; 2 2 [1 + (x â&#x2C6;&#x2019; 2) ][16 + (5 â&#x2C6;&#x2019; x) ] â&#x2C6;&#x161; â&#x2C6;&#x161; 2 Âą 4 + 28 = 1 Âą 2 2, dÎ¸/dx = 0 when x = 2 â&#x2C6;&#x161; â&#x2C6;&#x161; > 0 for x in [2, 1 + 2 2), only 1 + 2 2 is in [2, 5]; dÎ¸/dx â&#x2C6;&#x161; â&#x2C6;&#x161; dÎ¸/dx < 0 for x in (1 + 2 2, 5], Î¸ is maximum when x = 1 + 2 2. 2

62. Î¸ = Îą â&#x2C6;&#x2019; Î˛

= cotâ&#x2C6;&#x2019;1 (x/12) â&#x2C6;&#x2019; cotâ&#x2C6;&#x2019;1 (x/2)

10(24 â&#x2C6;&#x2019; x2 ) (144 + x2 )(4 + x2 ) â&#x2C6;&#x161; â&#x2C6;&#x161; dÎ¸/dx = 0 when x = 24 = 2 6, by the ďŹ rst derivative test Î¸ is maximum there.

12 2 dÎ¸ =â&#x2C6;&#x2019; + dx 144 + x2 4 + x2 =

â?Ş

â?Ł

â?¤

63. Let v = speed of light in the medium. The total time required for the light to travel from A to P to B is 1 t = (total distance from A to P to B)/v = ( (c â&#x2C6;&#x2019; x)2 + a2 + x2 + b2 ), v 1 câ&#x2C6;&#x2019;x x dt = â&#x2C6;&#x2019; +â&#x2C6;&#x161; 2 2 2 dx v x + b2 (c â&#x2C6;&#x2019; x) + a â&#x2C6;&#x161; dt câ&#x2C6;&#x2019;x x = . But x/ x2 + b2 = sin Î¸2 and = 0 when â&#x2C6;&#x161; dx x2 + b2 (c â&#x2C6;&#x2019; x)2 + a2 (c â&#x2C6;&#x2019; x)/ (c â&#x2C6;&#x2019; x)2 + a2 = sin Î¸1 thus dt/dx = 0 when sin Î¸2 = sin Î¸1 so Î¸2 = Î¸1 .

and

64. The total time required for the light to travel from A to P to B is â&#x2C6;&#x161; (c â&#x2C6;&#x2019; x)2 + b2 x2 + a2 + , t = (time from A to P ) + (time from P to B) = v1 v2 â&#x2C6;&#x161; x dt câ&#x2C6;&#x2019;x = â&#x2C6;&#x161; â&#x2C6;&#x2019; but x/ x2 + a2 = sin Î¸1 and dx v1 x2 + a2 v2 (c â&#x2C6;&#x2019; x)2 + b2

uh a

dt sin Î¸2 dt sin Î¸2 sin Î¸1 sin Î¸1 â&#x2C6;&#x2019; so = . (c â&#x2C6;&#x2019; x)/ (c â&#x2C6;&#x2019; x)2 + b2 = sin Î¸2 thus = = 0 when dx v1 v2 dx v1 v2

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Exercise Set 5.7

203

(b) the best path occurs when θ1 = θ2 (see ﬁgure).

House

θ2

θ1

us uf

Barn

3 4

1 4

1−x

EXERCISE SET 5.7 x2n − 2 2xn x1 = 1, x2 = 1.5, x3 = 1.416666667, . . . , x5 = x6 = 1.414213562

1. f (x) = x2 − 2, f (x) = 2x, xn+1 = xn −

x2n − 7 2xn x1 = 3, x2 = 2.666666667, x3 = 2.645833333, . . . , x5 = x6 = 2.645751311

2. f (x) = x2 − 7, f (x) = 2x, xn+1 = xn −

x3n − 6 3x2n x1 = 2, x2 = 1.833333333, x3 = 1.817263545, . . . , x5 = x6 = 1.817120593

3. f (x) = x3 − 6, f (x) = 3x2 , xn+1 = xn −

4. xn − a = 0

65. (a) The rate at which the farmer walks is analogous to the speed of light in Fermat’s principle.

x3n − xn + 3 3x2n − 1 x1 = −2, x2 = −1.727272727, x3 = −1.673691174, . . . , x5 = x6 = −1.671699882

5. f (x) = x3 − x + 3, f (x) = 3x2 − 1, xn+1 = xn −

x3n + xn − 1 3x2n + 1 x1 = 1, x2 = 0.75, x3 = 0.686046512, . . . , x5 = x6 = 0.682327804

6. f (x) = x3 + x − 1, f (x) = 3x2 + 1, xn+1 = xn −

x5n + x4n − 5 5x4n + 4x3n x1 = 1, x2 = 1.333333333, x3 = 1.239420573, . . . , x6 = x7 = 1.224439550

7. f (x) = x5 + x4 − 5, f (x) = 5x4 + 4x3 , xn+1 = xn −

x5n − xn + 1 5x4n − 1 x1 = −1, x2 = −1.25, x3 = −1.178459394, . . . , x6 = x7 = −1.167303978

uh a

8. f (x) = x5 − x + 1, f (x) = 5x4 − 1, xn+1 = xn −

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204

Chapter 5

9. f (x) = x4 + x − 3, f (x) = 4x3 + 1,

x1 = −2, x2 = −1.645161290, x3 = −1.485723955, . . . , x6 = x7 = −1.452626879

us uf

xn+1

x4 + xn − 3 = xn − n 3 4xn + 1 -2

-6

xn+1

x5 − 5x3n − 2 = xn − n 4 5xn − 15x2n

-2.5

11. f (x) = 2 sin x − x, f (x) = 2 cos x − 1, 2 sin xn − xn = xn − 2 cos xn − 1

ss Ha

12. f (x) = sin x − x2 , f (x) = cos x − 2x, sin xn − x2n = xn − cos xn − 2xn

2.5

-20

x1 = 2, x2 = 1.900995594, x3 = 1.895511645, x4 = x5 = 1.895494267

xn+1

x1 = 2, x2 = 2.5, x3 = 2.327384615, . . . , x7 = x8 = 2.273791732

xn+1

10. f (x) = x5 − 5x3 − 2, f (x) = 5x4 − 15x2 ,

–7

0.3 0

1.5

x1 = 1, x2 = 0.891395995, x3 = 0.876984845, . . . , x5 = x6 = 0.876726215 –1.3

13. f (x) = x − tan x,

f (x) = 1 − sec x = − tan x, xn+1

xn − tan xn = xn + tan2 xn

–100

uh a

x1 = 4.5, x2 = 4.493613903, x3 = 4.493409655, x4 = x5 = 4.493409458

100

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Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Exercise Set 5.7

205

14. f (x) = 1 + ex cos x, f (x) = ex (cos x − sin x) 3

1 + exn cos xn exn (cos xn − sin xn )

us uf

xn+1 = xn −

x1 = 2, x2 = 1.7881, x3 = 1.74757, x4 = 1.74614126, x5 = 1.746139530

xn+1 = xn −

-2

15. At the point of intersection, x3 = 0.5x − 1, x3 − 0.5x + 1 = 0. Let f (x) = x3 − 0.5x + 1. By graphing y = x3 and y = 0.5x − 1 it is evident that there is only one point of intersection and it occurs in the interval [−2, −1]; note that f (−2) < 0 and f (−1) > 0. f (x) = 3x2 − 0.5 so

–22

x3n − 0.5x + 1 ; 3x2n − 0.5

-2

x1 = −1, x2 = −1.2, x3 = −1.166492147, . . . , x5 = x6 = −1.165373043

16. The graphs of y = sin x and y = x3 − 2x2 + 1 intersect at points near x = −0.8 and x = 0.6 and x = 2. Let f (x) = sin x−x3 +2x2 −1, then f (x) = cos x−3x2 +4x, so

cos x − 3x2 + 4x . sin x − x3 + 2x2 + 1

xn+1 = xn −

2.5

-1.5

If x1 = −0.8, then x2 = −0.783124811, x3 = −0.782808234, x4 = x5 = −0.782808123; if x1 = 0.6, then x2 = 0.568003853, x3 = x4 = 0.568025739 ; if x1 = 2, then x2 = 1.979461151, x3 = 1.979019264, x4 = x5 = 1.979019061

–1

√ 17. The graphs of y = x2 and y = 2x + √1 intersect at points near x = −0.5 and x = 1; x2 = 2x + 1, x4 − 2x − 1 = 0. Let f (x) = x4 − 2x − 1, then f (x) = 4x3 − 2 so

x4n − 2xn − 1 . 4x3n − 2

xn+1 = xn −

-0.5

2 0

uh a

If x1 = −0.5, then x2 = −0.475, x3 = −0.474626695, x4 = x5 = −0.474626618; if x1 = 1, then x2 = 2, x3 = 1.633333333, . . . , x8 = x9 = 1.395336994.

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Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Chapter 5

us uf

xn+1 = xn −

18. The graphs of y = x3 /8 + 1 and y = cos 2x intersect at x = 0 and at a point near x = −2; x3 /8 + 1 = cos 2x, x3 − 8 cos 2x + 8 = 0. Let f (x) = x3 − 8 cos 2x + 8, then f (x) = 3x2 + 16 sin 2x so

-3

x3n − 8 cos 2xn + 8 . 3x2n + 16 sin 2xn

-2

x1 = −2, x2 = −2.216897577, x3 = −2.193821581, . . . , x5 = x6 = −2.193618950. 19. x = 0; also set f (x) = 1 − ex cos x, f (x) = ex (sin x − cos x), 1 − ex cos x ex (sin x − cos x)

xn+1 = xn −

206

x1 = 1, x2 = 1.572512605, x3 = 1.363631415, x7 = x8 = 1.292695719

-5

20. The graphs of y = e−x and y = ln x intersect near x = 1.3; let f (x) = e−x − ln x, f (x) = −e−x − 1/x, x1 = 1.3,

e−xn − ln xn , x2 = 1.309759929, e−xn + 1/xn x4 = x5 = 1.309799586

xn+1 = xn +

21. (a) f (x) = x − a, f (x) = 2x, xn+1 2

x2 − a 1 = xn − n = 2xn 2

-4

a xn + xn

22. (a) f (x) =

(b) a = 10; x1 = 3, x2 = 3.166666667, x3 = 3.162280702, x4 = x5 = 3.162277660 1 1 − a, f (x) = − 2 , xn+1 = xn (2 − axn ) x x

(b) a = 17; x1 = 0.05, x2 = 0.0575, x3 = 0.058793750, x5 = x6 = 0.058823529

uh a

23. f (x) = x3 + 2x + 5; solve f (x) = 0 to ﬁnd the critical points. Graph y = x3 and y = −2x − 5 to x3 + 2xn + 5 see that they intersect at a point near x = −1; f (x) = 3x2 + 2 so xn+1 = xn − n 2 . 3xn + 2 x1 = −1, x2 = −1.4, x3 = −1.330964467, · · · , x5 = x6 = −1.328268856 so the minimum value of f (x) occurs at x ≈ −1.328268856 because f (x) > 0; its value is approximately −4.098859132.

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Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Exercise Set 5.7

207

24. From a rough sketch of y = x sin x we see that the maximum occurs at a point near x = 2, which will be a point where f (x) = x cos x + sin x = 0. f (x) = 2 cos x − x sin x so

xn cos xn + sin xn xn + tan xn = xn − . 2 cos xn − xn sin xn 2 − xn tan xn

us uf

xn+1 = xn −

x1 = 2, x2 = 2.029048281, x3 = 2.028757866, x4 = x5 = 2.028757838; the maximum value is approximately 1.819705741.

25. A graphing utility shows that there are two inﬂection points at x ≈ −0.25, 1.25. These points ex are the zeros of f (x) = (x4 − 4x3 + 8x2 − 4x − 1) 2 . It is equivalent to ﬁnd the zeros of (x + 1)3 g(x) = x4 − 4x3 + 8x2 − 4x − 1. One root is x = 1 by inspection. Since g (x) = 4x3 − 12x2 + 16x − 4, Newton’s Method becomes x4n − 4x3n + 8x2n − 4xn − 1 4x3n − 12x2n + 16xn − 4

xn+1 = xn −

With x0 = −0.25, x1 = −0.18572695, x2 = −0.179563312, x3 = −0.179509029, x4 = x5 = −0.179509025. So the points of inﬂection are at x ≈ −0.18, x = 1. 1 − 2x = 0 for x = x1 ≈ 0.2451467013, f (x1 ) ≈ 0.1225363521 x2 + 1

26. f (x) = −2 tan−1 x +

27. Let f (x) be the square of the distance between (1, 0) and any point (x, x2 ) on the parabola, then f (x) = (x − 1)2 + (x2 − 0)2 = x4 + x2 − 2x + 1 and f (x) = 4x3 + 2x − 2. Solve f (x) = 0 to ﬁnd 4x3 + 2xn − 2 2x3 + xn − 1 = xn − n 2 . the critical points; f (x) = 12x2 + 2 so xn+1 = xn − n 2 12xn + 2 6xn + 1 x1 = 1, x2 = 0.714285714, x3 = 0.605168701, . . . , x6 = x7 = 0.589754512; the coordinates are approximately (0.589754512, 0.347810385).

28. The area is A = xy = x cos x so dA/dx = cos x − x sin x. Find x so that dA/dx = 0; d2 A/dx2 = −2 sin x − x cos x so xn+1 = xn +

cos xn − xn sin xn 1 − xn tan xn = xn + . 2 sin xn + xn cos xn 2 tan xn + xn

x1 = 1, x2 = 0.864536397, x3 = 0.860339078, x4 = x5 = 0.860333589; y ≈ 0.652184624.

29. (a) Let s be the arc length, and L the length of the chord, then s = 1.5L. But s = rθ and L = 2r sin(θ/2) so rθ = 3r sin(θ/2), θ − 3 sin(θ/2) = 0. θn − 3 sin(θn /2) (b) Let f (θ) = θ − 3 sin(θ/2), then f (θ) = 1 − 1.5 cos(θ/2) so θn+1 = θn − . 1 − 1.5 cos(θn /2) θ1 = 3, θ2 = 2.991592920, θ3 = 2.991563137, θ4 = θ5 = 2.991563136 rad so θ ≈ 171◦ .

uh a

30. r2 (θ − sin θ)/2 = πr2 /4 so θ − sin θ − π/2 = 0. Let f (θ) = θ − sin θ − π/2, then f (θ) = 1 − cos θ θn − sin θn − π/2 so θn+1 = . 1 − cos θn θ1 = 2, θ2 = 2.339014106, θ3 = 2.310063197, . . . , θ5 = θ6 = 2.309881460 rad; θ ≈ 132◦ .

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208

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Chapter 5

us uf

If y1 = −1, then y2 = −1.333333333, y3 = −1.235807860, . . . , y6 = y7 = −1.220744085; if y1 = 1, then y2 = 0.8, y3 = 0.731233596, . . . , y6 = y7 = 0.724491959.

31. If x = 1, then y 4 + y = 1, y 4 + y − 1 = 0. Graph z = y 4 and z = 1 − y to see that they intersect y 4 + yn − 1 . near y = −1 and y = 1. Let f (y) = y 4 + y − 1, then f (y) = 4y 3 + 1 so yn+1 = yn − n 3 4yn + 1

32. If x = 1, then 2y − cos y = 0. Graph z = 2y and z = cos y to see that they intersect near y = 0.5. 2yn − cos yn Let f (y) = 2y − cos y, then f (y) = 2 + sin y so yn+1 = yn − . 2 + sin yn y1 = 0.5, y2 = 0.450626693, y3 = 0.450183648, y4 = y5 = 0.450183611. 5000 (1 + i)25 − 1 ; set f (i) = 50i − (1 + i)25 + 1, f (i) = 50 − 25(1 + i)24 ; solve i f (i) = 0. Set i0 = .06 and ik+1 = ik − 50i − (1 + i)25 + 1 / 50 − 25(1 + i)24 . Then i1 = 0.05430, i2 = 0.05338, i3 = 0.05336, . . . , i = 0.053362.

34. (a) x1 = 2, x2 = 5.3333, x3 = 11.055, x4 = 22.293, x5 = 44.676

33. S(25) = 250,000 =

0.5

0.5000

−0.7500

0.2917

−1.5685

x10

−0.4654

0.8415

−0.1734

2.7970

1.2197

0.1999

35. (a)

(b) x1 = 0.5, x2 = −0.3333, x3 = 0.0833, x4 = −0.0012, x5 = 0.0000 (and xn = 0 for n ≥ 6)

(b) The sequence xn must diverge, since if it did converge then f (x) = x2 + 1 = 0 would have a solution. It seems the xn are oscillating back and forth in a quasi-cyclical fashion.

EXERCISE SET 5.8

1. f (0) = f (4) = 0; f (3) = 0; [0, 4], c = 3

2. f (−3) = f (3) = 0; f (0) = 0

uh a

3. f (2) = f (4) = 0, f (x) = 2x − 6, 2c − 6 = 0, c = 3

4. f (0) = f (2) = 0, f (x) = 3x − 6x + 2, 3c − 6c + 2 = 0; c = 2

6±

√

√ 36 − 24 = 1 ± 3/3 6

5. f (π/2) = f (3π/2) = 0, f (x) = − sin x, − sin c = 0, c = π

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Exercise Set 5.8

x2 − 4x + 1 c2 − 4c + 1 , = 0, c2 − 4c + 1 = 0 (x − 2)2 (c − 2)2

6. f (−1) = f (1) = 0, f (x) = √

1 1 1 1 − √ , − √ = 0, c = 1 2 2 x 2 2 c

8. f (1) = f (3) = 0, f (x) = −

3 6 f (8) − f (0) = = = f (1.54); c = 1.54 8 4 8−0 f (4) − f (0) = 1.19 = f (0.77) 4−0

11. f (−4) = 12, f (6) = 42, f (x) = 2x + 1, 2c + 1 =

42 − 12 = 3, c = 1 6 − (−4)

10.

2 4 2 4 + 2 , − 3 + 2 = 0, −6 + 4c = 0, c = 3/2 3 x 3x c 3c

12. f (−1) = −6, f (2) = 6, f (x) = 3x2 + 1, 3c2 + 1 = c = 1 is in (−1, 2)

6 − (−6) = 4, c2 = 1, c = ±1 of which only 2 − (−1)

us uf

7. f (0) = f (4) = 0, f (x) =

√ √ 16 − 4 = 2 ± 3, of which only c = 2 − 3 is in (−1, 1) 2

4±

209

1 √ 1 1 2−1 = , c + 1 = 3/2, c + 1 = 9/4, c = 5/4 13. f (0) = 1, f (3) = 2, f (x) = √ , √ = 3−0 3 2 x+1 2 c+1

14. f (3) = 10/3, f (4) = 17/4, f (x) = 1 − 1/x2 , 1 − 1/c2 = √ of which only c = 2 3 is in (3, 4)

√ 17/4 − 10/3 = 11/12, c2 = 12, c = ±2 3 4−3

√ x c 4−0 1 , −√ = 15. f (−5) = 0, f (3) = 4, f (x) = − √ = , −2c = 25 − c2 , 2 2 3 − (−5) 2 25 − x 25 − c √ c2 = 5, c = − 5 4c2 = 25 − c2 , √ √ (we reject c = 5 because it does not satisfy the equation −2c = 25 − c2 ) 16. f (2) = 1, f (5) = 1/4, f (x) = −1/(x − 1)2 , −

c = −1 (reject), or c = 3

(b) c = −1.29

uh a

17. (a) f (−2) = f (1) = 0 The interval is [−2, 1]

1 1/4 − 1 1 = = − , (c − 1)2 = 4, c − 1 = ±2, (c − 1)2 5−2 4

-2

1 -2

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210

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Chapter 5

18. (a) m =

−16 − 5 f (−2) − f (1) = = 7 so y − 5 = 7(x − 1), y = 7x − 2 −2 − 1 −3

(d)

us uf

-2

-20

20. (a) f (−1) = 1, f (8) = 4, f (x) =

(b) tan x is not continuous on [0, π]

19. (a) f (x) = sec2 x, sec2 c = 0 has no solution

(b) f (x) = 3x2 + 4 = 7 has solutions x = ±1; discard x = 1, so c = −1

2 −1/3 x 3

1 2 −1/3 4−1 = , c1/3 = 2, c = 8 which is not in (−1, 8). = c 3 8 − (−1) 3 (b) x2/3 is not diﬀerentiable at x = 0, which is in (−1, 8).

21. (a) Two x-intercepts of f determine two solutions a and b of f (x) = 0; by Rolle’s Theorem there exists a point c between a and b such that f (c) = 0, i.e. c is an x-intercept for f .

f (x1 ) − f (x0 ) is the average rate of change of y with respect to x on the interval [x0 , x1 ]. By x1 − x0 the Mean-Value Theorem there is a value c in (x0 , x1 ) such that the instantaneous rate of change f (x1 ) − f (x0 ) f (c) = . x1 − x0

22.

(b) f (x) = sin x = 0 at x = nπ, and f (x) = cos x = 0 at x = nπ + π/2, which lies between nπ and (n + 1)π, (n = 0, ±1, ±2, . . .)

23. Let s(t) be the position function of the automobile for 0 ≤ t ≤ 5, then by the Mean-Value Theorem there is at least one point c in (0, 5) where s (c) = v(c) = [s(5) − s(0)]/(5 − 0) = 4/5 = 0.8 mi/min = 48 mi/h.

24. Let T (t) denote the temperature at time with t = 0 denoting 11 AM, then T (0) = 76 and T (12) = 52.

uh a

(a) By the Mean-Value Theorem there is a value c between 0 and 12 such that T (c) = [T (12) − T (0)]/(12 − 0) = (52 − 76)/(12) = −2◦ F/h. (b) Assume that T (t1 ) = 88◦ F where 0 < t1 < 12, then there is at least one point c in (t1 , 12) where T (c) = [T (12)−T (t1 )]/(12−t1 ) = (52−88)/(12−t1 ) = −36/(12−t1 ). But 12−t1 < 12 so T (c) < −36/12 = −3◦ F/h.

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Exercise Set 5.8

211

us uf

25. Let f (t) and g(t) denote the distances from the first and second runners to the starting point, and let h(t) = f (t) − g(t). Since they start (at t = 0) and finish (at t = t1 ) at the same time, h(0) = h(t1 ) = 0, so by Rolle’s Theorem there is a time t2 for which h (t2 ) = 0, i.e. f (t2 ) = g (t2 ); so they have the same velocity at time t2 . 26. f (x) = x6 − 2x2 + x satisfies f (0) = f (1) = 0, so by Rolle’s Theorem f (c) = 0 for some c in (0, 1).

27. (a) By the Constant Diﬀerence Theorem f (x) − g(x) = k for some k; since f (x0 ) = g(x0 ), k = 0, so f (x) = g(x) for all x.

(b) Set f (x) = sin2 x + cos2 x, g(x) = 1; then f (x) = 2 sin x cos x − 2 cos x sin x = 0 = g (x). Since f (0) = 1 = g(0), f (x) = g(x) for all x.

28. (a) By the Constant Diﬀerence Theorem f (x) − g(x) = k for some k; since f (x0 ) − g(x0 ) = c, k = c, so f (x) − g(x) = c for all x.

(b) Set f (x) = (x − 1)3 , g(x) = (x2 + 3)(x − 3). Then f (x) = 3(x − 1)2 , g (x) = (x2 + 3) + 2x(x − 3) = 3x2 − 6x + 3 = 3(x2 − 2x + 1) = 3(x − 1)2 , so f (x) = g (x) and hence f (x) − g(x) = k. Expand f (x) and g(x) to get h(x) = f (x) − g(x) = (x3 − 3x2 + 3x − 1) − (x3 − 3x2 + 3x − 9) = 8.

29. If f (x) = g (x), then f (x) = g(x) + k. Let x = 1, f (1) = g(1) + k = (1)3 − 4(1) + 6 + k = 3 + k = 2, so k = −1. f (x) = x3 − 4x + 5. 30. By the Constant Diﬀerence Theorem f (x) = tan−1 x + C and

2 = f (1) = tan−1 (1) + C = π/4 + C, C = 2 − π/4, f (x) = tan−1 x + 2 − π/4. f (y) − f (x) = f (c), y−x so |f (x) − f (y)| = |f (c)||x − y| ≤ M |x − y|; if x > y exchange x and y; if x = y the inequality also holds.

31. (a) If x, y belong to I and x < y then for some c in I,

(b) f (x) = sin x, f (x) = cos x, |f (x)| ≤ 1 = M , so |f (x) − f (y)| ≤ |x − y| or | sin x − sin y| ≤ |x − y|. f (y) − f (x) = f (c), y−x so |f (x) − f (y)| = |f (c)||x − y| ≥ M |x − y|; if x > y exchange x and y; if x = y the inequality also holds.

32. (a) If x, y belong to I and x < y then for some c in I,

(b) If x and y belong to (−π/2, π/2) and f (x) = tan x, then |f (x)| = sec2 x ≥ 1 and | tan x − tan y| ≥ |x − y| (c) y lies in (−π/2, π/2) if and only if −y does; use Part (b) and replace y with −y

uh a

√ 33. (a) Let f (x) √ = x. By the Mean-Value Theorem there is a number c between x and y such that √ √ y− x 1 1 y−x √ = √ < √ for c in (x, y), thus y − x < √ y−x 2 c 2 x 2 x 1 √ (b) multiply through and rearrange to get xy < (x + y). 2

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212

Chapter 5

us uf

34. Suppose that f (x) has at least two distinct real solutions r1 and r2 in I. Then f (r1 ) = f (r2 ) = 0 so by Rolle’s Theorem there is at least one number between r1 and r2 where f (x) = 0, but this contradicts the assumption that f (x) = 0, so f (x) = 0 must have fewer than two distinct solutions in I.

35. (a) If f (x) = x3 + 4x − 1 then f (x) = 3x2 + 4 is never zero, so by Exercise 34 f has at most one real root; since f is a cubic polynomial it has at least one real root, so it has exactly one real root.

(b) Let f (x) = ax3 + bx2 + cx + d. If f (x) = 0 has at least two distinct real solutions r1 and r2 , then f (r1 ) = f (r2 ) = 0 and by Rolle’s Theorem there is at least one number between r1 and 2 r2 where f (x) 2bx + c = 0 for √ = 0. But f (x) = 3ax +√ 2 x = (−2b ± 4b − 12ac)/(6a) = (−b ± b2 − 3ac)/(3a), which are not real if b2 − 3ac < 0 so f (x) = 0 must have fewer than two distinct real solutions. √ √ √ 1 1 1 4− 3 1 1 = 2 − 3. But < √ < √ for c in (3, 4) so 36. f (x) = √ , √ = 4−3 4 2 c 2 x 2 c 2 3 √ √ √ √ 1 1 < 2 − 3 < √ , 0.25 < 2 − 3 < 0.29, −1.75 < − 3 < −1.71, 1.71 < 3 < 1.75. 4 2 3

37. By the Mean-Value Theorem on the interval [0, x],

d 2 [f (x) − g 2 (x)] = 2f (x)f (x) − 2g(x)g (x) = 2f (x)g(x) − 2g(x)f (x) = 0, so f 2 − g 2 is dx constant.

38. (a)

tan−1 x 1 tan−1 x − tan−1 0 = = for c in (0, x), but x−0 x 1 + c2 1 1 tan−1 x 1 x < < 1 for c in (0, x) so < < tan−1 x < x. < 1, 2 2 2 1+x 1+c 1+x x 1 + x2

(b) f (x) = 12 (ex − e−x ) = g(x), g (x) = 12 (ex + e−x ) = f (x) d 2 [f (x) + g 2 (x)] = 2f (x)f (x) + 2g(x)g (x) = 2f (x)g(x) + 2g(x)[−f (x)] = 0, dx so f 2 (x) + g 2 (x) is constant.

39. (a)

(b) f (x) = sin x and g(x) = cos x

]

uh a

41.

40. Let h = f − g, then h is continuous on [a, b], diﬀerentiable on (a, b), and h(a) = f (a) − g(a) = 0, h(b) = f (b) − g(b) = 0. By Rolle’s Theorem there is some c in (a, b) where h (c) = 0. But h (c) = f (c) − g (c) so f (c) − g (c) = 0, f (c) = g (c).

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Exercise Set 5.8

213

42. (a) Suppose f (x) = 0 more than once in (a, b), say at c1 and c2 . Then f (c1 ) = f (c2 ) = 0 and by using Rolle’s Theorem on f , there is some c between c1 and c2 where f (c) = 0, which contradicts the fact that f (x) > 0 so f (x) = 0 at most once in (a, b).

us uf

(b) If f (x) > 0 for all x in (a, b), then f is concave up on (a, b) and has at most one relative extremum, which would be a relative minimum, on (a, b). 43. (a) similar to the proof of Part (a) with f (c) < 0 (b) similar to the proof of Part (a) with f (c) = 0

44. Let x = x0 be suﬃciently near x0 so that there exists (by the Mean-Value Theorem) a number c (which depends on x) between x and x0 , such that f (x) − f (x0 = f (c). x − x0

x→x0

f (x) − f (x0 ) x − x0

(by deﬁnition of derivative)

f (x0 ) = lim

Since c is between x and x0 it follows that

= lim f (c)

(by the Mean-Value Theorem)

= lim f (x)

(since lim f (x) exists and c is between x and x0 ).

x→x0

45. If f is diﬀerentiable at x = 1, then f is continuous there;

lim f (x) = lim f (x) = f (1) = 3, a + b = 3; lim f (x) = a and x→1−

x→1+

lim f (x) = 6 so a = 6 and b = 3 − 6 = −3.

x→1−

lim f (x) = lim− 2x = 0 and lim+ f (x) = lim+ 2x = 0; f (0) does not exist because f is

x→0−

x→0

46. (a)

not continuous at x = 0.

(b)

x→0

lim f (x) = lim+ f (x) = 0 and f is continuous at x = 0, so f (0) = 0;

x→0−

x→0

lim− f (x) = lim− (2) = 2 and lim+ f (x) = lim+ 6x = 0, so f (0) does not exist. x→0

x→0

x→x0

47. From Section 3.2 a function has a vertical tangent line at a point of its graph if the slopes of secant lines through the point approach +∞ or −∞. Suppose f is continuous at x = x0 and lim f (x) = +∞. Then a secant line through (x1 , f (x1 )) and (x0 , f (x0 )), assuming x1 > x0 , will + f (x1 ) − f (x0 ) . By the Mean Value Theorem, this quotient is equal to f (c) for some x1 − x0

have slope

c between x0 and x1 . But as x1 approaches x0 , c must also approach x0 , and it is given that lim+ f (c) = +∞, so the slope of the secant line approaches +∞. The argument can be altered

c→x0

uh a

appropriately for x1 < x0 , and/or for f (c) approaching −∞.

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214

Chapter 5

SUPPLEMENTARY EXERCISES FOR CHAPTER 5 x2 x3 − on [−2, 2]; x = 0 is a relative maximum and x = 1 is a 3 2 1 relative minimum, but y = 0 is not the largest value of y on the interval, nor is y = − the 6 smallest. (b) true (c) False; for example y = x3 on (−1, 1) which has a critical number but no relative extrema

(c)

us uf

(b)

x 2

6. (a)

4. (a) False; an example is y =

7(x − 7)(x − 1) ; critical numbers at x = 0, 1, 7; 3x2/3 neither at x = 0, relative maximum at x = 1, relative minimum at x = 7 (First Derivative Test)

7. (a) f (x) =

(b) f (x) = 2 cos x(1 + 2 sin x); critical numbers at x = π/2, 3π/2, 7π/6, 11π/6; relative maximum at x = π/2, 3π/2, relative minimum at x = 7π/6, 11π/6 √ 3 x−1 ; critical numbers at x = 5; relative maximum at x = 5 (c) f (x) = 3 − 2 x−9 27 − x , f (x) = ; critical number at x = 9; 18x3/2 36x5/2 f (9) > 0, relative minimum at x = 9

8. (a) f (x) =

x3 − 4 x3 + 8 , f (x) = 2 ; 2 x x3 critical number at x = 41/3 , f (41/3 ) > 0, relative min at x = 41/3

(b) f (x) = 2

(c) f (x) = sin x(2 cos x + 1), f (x) = 2 cos2 x − 2 sin2 x + cos x; critical numbers at x = 2π/3, π, 4π/3; f (2π/3) < 0, relative maximum at x = 2π/3; f (π) > 0, relative minimum at x = π; f (4π/3) < 0, relative maximum at x = 4π/3 y

lim f (x) = +∞, lim f (x) = +∞

x→−∞

x→+∞

f (x) = x(4x − 9x + 6), f (x) = 6(2x − 1)(x − 1) relative minimum at x = 0, points of inﬂection when x = 1/2, 1, no asymptotes

3 2 1

(1,2)

(12 ,2316)

(0,1)

x 1

uh a

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Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Supplementary Exercises for Chapter 5

lim f (x) doesn’t exist

x→±∞

f (x) = 2x sec2 (x2 + 1), f (x) = 2 sec2 (x2 + 1) 1 + 4x2 tan(x2 + 1) critical number at x = 0; relative minimum at x = 0

-100 -200

point of inﬂection when 1 + 4x2 tan(x2 + 1) = 0 vertical asymptotes at x = ± π(n + 21 ) − 1, n = 0, 1, 2, . . .

y 4 2 x

-2

-1

x→+∞

-4

f (x) = 1 + sin x, f (x) = cos x critical numbers at x = 2nπ + π/2, n = 0, ±1, ±2, . . .,

no extrema because f ≥ 0 and by Exercise 53 of Section 5.1, f is increasing on (−∞, +∞)

-c

-6

x(x + 5) 2x3 + 15x2 − 25 , f (x) = −2 (x2 + 2x + 5)2 (x2 + 2x + 5)3 critical numbers at x = −5, 0; relative maximum at x = −5, relative minimum at x = 0 points of inﬂection at x = −7.26, −1.44, 1.20 horizontal asymptote y = 1 as x → ±∞

1 0.8 0.6 0.4 0.2 x -20

-10

uh a

c -2 -4

inﬂections points at x = nπ + π/2, n = 0, ±1, ±2, . . . no asymptotes

13. f (x) = 2

-2

lim f (x) = −∞, lim f (x) = +∞

x→−∞

(3.83,-261.31)

12.

(2.82,-165.00)

11.

(0,0) 1 2 3 4

f (x) = x3 (x − 2)2 , f (x) = x2 (5x − 6)(x − 2), f (x) = 4x(5x2 − 12x + 6) √ 8 ± 2 31 critical numbers at x = 0, 5 √ 8 − 2 31 ≈ −0.63 relative maximum at x = 5 √ 8 + 2 31 relative minimum at x = ≈ 3.83 5 √ 6 ± 66 points of inﬂection at x = 0, ≈ 0, −0.42, 2.82 5 no asymptotes

(-0.42,0.16) (-0.63,0.27) -1

x→+∞

us uf

lim f (x) = −∞, lim f (x) = +∞

x→−∞

10.

215

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Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Chapter 5

3x2 − 25 3x2 − 50 , f (x) = −6 4 x x5 √ critical numbers at x = ±5 3/3; √ relative maximum at x = −5 3/3, √ relative minimum at x = +5 3/3 inﬂection points at x = ±5 2/3 horizontal asymptote of y = 0 as x → ±∞, vertical asymptote x = 0

14. f (x) = 3

-5

lim f (x) = +∞, lim f (x) = −∞ x→−∞ x→+∞ x x≤0 f (x) = if −2x x>0

2 1 -2

critical number at x = 0, no extrema inﬂection point at x = 0 (f changes concavity) no asymptotes

-2

−32 9(1 + x)4/3 (3 − x)5/3

y 4 2

critical number at x = 5/3; relative maximum at x = 5/3

x -4

-2

-1

-3

17. (a)

-5

cusp at x = −1; point of inﬂection at x = 3 oblique asymptote y = −x as x → ±∞

f (x) =

5 − 3x , 3(1 + x)1/3 (3 − x)2/3

16. f (x) =

x -4

15.

us uf

216

-40

1 , f (x) = 2x 400 1 critical points at x = ± ; 20 1 relative maximum at x = − , 20 1 relative minimum at x = 20

(b) f (x) = x2 −

0.0001

-0.1

0.1

-0.0001

uh a

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Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Supplementary Exercises for Chapter 5

-5

-200

(c)

-2.909

-2.2

3.5 1.3 -2.912

1.6

-4

19. (a)

√ (b) critical points at x = ± 2, 32 , 2; √ relative maximum at x = − 2, √ relative minimum at x = 2, relative maximum at x = 32 , relative minimum at x = 2

200

us uf

18. (a)

217

(b) Divide y = x2 + 1 into y = x3 − 8 to get the asymptote ax + b = x 5

-5

20. (a) p(x) = x3 − x (c) p(x) = x5 − x4 − x3 + x2

-6

(b) p(x) = x4 − x2 (d) p(x) = x5 − x3

21. f (x) = 4x3 − 18x2 + 24x − 8, f (x) = 12(x − 1)(x − 2) f (1) = 0, f (1) = 2, f (1) = 2; f (2) = 0, f (2) = 0, f (2) = 3, so the tangent lines at the inﬂection points are y = 2x and y = 3. dy dy dy dy cos x =2 ; = 0 when cos x = 0. Use the ﬁrst derivative test: = dx dx dx dx 2 + sin y and 2 + sin y > 0, so critical points when cos x = 0, relative maxima when x = 2nπ + π/2, relative minima when x = 2nπ − π/2, n = 0, ±1, ±2, . . . (2x − 1)(x2 + x − 7) x2 + x − 7 = 2 , 2 (2x − 1)(3x + x − 1) 3x + x − 1

23. f (x) =

22. cos x − (sin y)

x = 1/2

5 x -4

2 -5

uh a

horizontal asymptote: y = 1/3, √ vertical asymptotes: x = (−1 ± 13)/6

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Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Chapter 5

(x − 2)(x2 + x + 1)(x2 − 2) (x − 2)(x2 − 2)2 (x2 + 1) x2 + x + 1 = 2 (x − 2)(x2 + 1)

(b)

f (x) =

24. (a)

us uf

218

x x=–¤

–1

x=¤

25. f (x) = 2ax + b; f (x) > 0 or f (x) < 0 on [0, +∞) if f (x) = 0 has no positive solution, so the polynomial is always increasing or always decreasing on [0, +∞) provided −b/2a ≤ 0.

26. f (x) = 3ax2 + 2bx + c; f (x) > 0 or f (x) < 0 on (−∞, +∞) if f (x) = 0 has no real solutions so from the quadratic formula (2b)2 − 4(3a)c < 0, 4b2 − 12ac < 0, b2 − 3ac < 0. If b2 − 3ac = 0, then f (x) = 0 has only one real solution at, say, x = c so f is always increasing or always decreasing on both (−∞, c] and [c, +∞), and hence on (−∞, +∞) because f is continuous everywhere. Thus f is always increasing or decreasing if b2 − 3ac ≤ 0. 2

27. (a) relative minimum −0.232466 at x = 0.450184

-1

1.5 -0.5

(b) relative maximum 0 at x = 0; relative minimum −0.107587 at x = ±0.674841

0.2

-1.2

1.2

-0.15

-1.5

1.5

-0.4

uh a

28. f (x) = 2 + 3x2 − 4x3 has one real root at x0 ≈ 1.136861168, so f (x) is increasing on (−∞, k)) at least for k = x0 . But f (x0 ) < 0, so f (x) has a relative maximum at x = x0 , and is thus decreasing to the right of x = x0 . So f is increasing on (−∞, x0 ], where x0 ≈ 1.136861.

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Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Supplementary Exercises for Chapter 5

29. (a) If a = k, a constant, then v = kt + b where b is constant; so the velocity changes sign at t = −b/k.

219

us uf

- b/ k b

(b) Consider the equation s = 5 − t3 /6, v = −t2 /2, a = −t. Then for t > 0, a is decreasing and av > 0, so the particle is speeding up.

30. s(t) = t/(t2 + 5), v(t) = (5 − t2 )/(t2 + 5)2 , a(t) = 2t(t2 − 15)/(t2 + 5)3 0.2

0.25

0.01

(a)

0 0

20 0

-0.05

s(t)

v(t)

20 -0.15

a(t)

√ (b) v changes sign at t = 5 √ √ (c) s = 5/10, v = 0, a = − 5/50 √ √ √ (d) a changes sign at √ t = 15,√so the particle is speeding up for 5 < t < 15, and it is slowing down for 0 < t < 5 and 15 < t √ (e) v(0) = 1/5, lim v(t) = 0, v(t) has one t-intercept at t = 5 and v(t) has one critical t→+∞ √ the maximum velocity occurs when t = 0 and the minimum point at t = 15. Consequently √ velocity occurs when t = 15.

uh a

31. (a) s(t) = s0 + v0 t − 12 gt2 = v0 t − 4.9t2 , v(t) = v0 − 9.8t; smax occurs when v = 0, i.e. t = v0 /9.8, √ and then 0.76 = smax = v0 (v0 /9.8) − 4.9(v0 /9.8)2 = v02 /19.6, so v0 = 0.76 · 19.6 = 3.86 m/s and s(t) = 3.86t − 4.9t2 . Then s(t) = 0 when t = 0, 0.7878, s(t) = 0.15 when t = 0.0410, 0.7468, and s(t) = 0.76 − 0.15 = 0.61 when t = 0.2188, 0.5689, so the player spends 0.5689 − 0.2188 = 0.3501 s in the top 15.0 cm of the jump and 0.0410 + (0.7878 − 0.7468) = 0.0820 s in the bottom 15.0 cm. (b) The height vs time plot is a parabola that opens down, and the slope is smallest near the top of the parabola, so a given change ∆h in height corresponds to a large time change ∆t near the top of the parabola and a narrower time change at points farther away from the top.

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220

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Chapter 5

32. (a) s(t) = s0 + v0 − 4.9t2 ; assume s0 = v0 = 0, so s(t) = −4.9t2 , v(t) = −9.8t 1 2 3 4 −4.9 −19.6 −44.1 −78.4 −9.8 −19.6 −29.4 −39.2

0 0 0

us uf

t s v

33. (a) v = −2

t(t4 + 2t2 − 1) 3t8 + 10t6 − 12t4 − 6t2 + 1 , a = 2 (t4 + 1)2 (t4 + 1)3

(b)

0.2

-0.2

(b) The formula for v is linear (with no constant term). (c) The formula for s is quadratic (with no linear or constant term).

t 2

(c) It is farthest from the origin at approximately t = 0.64 (when v = 0) and s = 1.2 (d) Find t so that the velocity v = ds/dt > 0. The particle is moving in the positive direction for 0 ≤ t ≤ 0.64 s. (e) It is speeding up when a, v > 0 or a, v < 0, so for 0 ≤ t < 0.36 and 0.64 < t < 1.1, otherwise it is slowing down. (f ) Find the maximum value of |v| to obtain: maximum speed = 1.05 m/s when t = 1.10 s.

34. No; speeding up means the velocity and acceleration have the same sign, i.e. av > 0; the velocity is increasing when the acceleration is positive, i.e. a > 0. These are not the same thing. An example is s = t − t2 at t = 1, where v = −1 and a = −2, so av > 0 but a < 0.

37. (a) If f has an absolute extremum at a point of (a, b) then it must, by Theorem 5.5.4, be at a critical point of f ; since f is diﬀerentiable on (a, b) the critical point is a stationary point. (b) It could occur at a critical point which is not a stationary point: for example, f (x) = |x| on [−1, 1] has an absolute minimum at x = 0 but is not diﬀerentiable there.

38. Yes; by the Mean-Value Theorem there is a point c in (a, b) such that f (c) =

f (b) − f (a) = 0. b−a

uh a

39. (a) f (x) = −1/x2 = 0, no critical points; by inspection M = −1/2 at x = −2; m = −1 at x = −1 (b) f (x) = 3x2 − 4x3 = 0 at x = 0, 3/4; f (−1) = −2, f (0) = 0, f (3/4) = 27/256, f (3/2) = −27/16, so m = −2 at x = −1, M = 27/256 at x = 3/4 1/3 2 x(7x − 12) 144 − (c) f (x) = , critical points at x = 12/7, 2; m = f (12/7) = ≈ −1.9356 49 7 3(x − 2)2/3 at x = 12/7, M = 9 at x = 3 ex (x − 2) (d) lim f (x) = lim f (x) = +∞ and f (x) = , stationary point at x = 2; by x→+∞ x3 x→0+ Theorem 5.5.5 f (x) has an absolute minimum at x = 2, and m = e2 /4.

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Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Supplementary Exercises for Chapter 5

221

√ 3 − x2 , critical point at x = 3. Since lim+ f (x) = 0, f (x) has no minimum, 2 2 (x + 3) x→0 √ √ and M = 3/3 at x = 3.

40. (a) f (x) = 2

us uf

(b) f (x) = 10x3 (x − 2), critical points at x = 0, 2; lim f (x) = 88, so f (x) has no maximum; x→3−

m = −9 at x = 2

(d) f (x) = (1 + ln x)xx , critical point at x = 1/e; lim f (x) = lim ex ln x = 1, x→0+

x→0+

42. x = −2.11491, 0.25410, 1.86081

43. x = 2.3561945

lim f (x) = +∞; no absolute maximum, absolute minimum m = e−1/e at x = 1/e

x→+∞

f (x) − f (x0 ) > 0 since f is increasing on [a, b]. Similarly if x0 < x < b x − x0 f (x) − f (x0 ) f (x) − f (x0 ) ≥ 0. > 0. Thus, since the limit exists, lim then x→x0 x − x0 x − x0

44. Let a < x < x0 . Then

45. (a) yes; f (0) = 0

46. (a) no, f is not diﬀerentiable on (−2, 2) √ f (3) − f (2) = −1 = f (1 + 2) 3−2

(b) yes,

lim f (x) = 2, lim+ f (x) = 2 so f is continuous on [0, 2]; lim− f (x) = lim− −2x = −2 and

x→1−

(c)

(b) no, f is not diﬀerentiable on (−1, 1) (c) yes, f ( π/2) = 0

x→1

lim f (x) = lim (−2/x2 ) = −2, so f is diﬀerentiable on (0, 2); + +

x→1

√ f (2) − f (0) = −1 = f ( 2). and 2−0

47. Let k be the amount of light admitted per unit area of clear glass. The total amount of light admitted by the entire window is 1 1 T = k · (area of clear glass) + k · (area of blue glass) = 2krh + πkr2 . 2 4 But P = 2h + 2r + πr which gives 2h = P − 2r − πr so

uh a

1 π 2 r T = kr(P − 2r − πr) + πkr2 = k P r − 2 + π − 4 4 8 + 3π 2 P r , = k Pr − for 0 < r < 4 2+π dT 8 + 3π dT 2P =k P − r , = 0 when r = . dr 2 dr 8 + 3π

This is the only critical point and d2 T /dr2 < 0 there so the most light is admitted when r = 2P/(8 + 3π) ft.

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222

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Chapter 5

49. (a)

us uf

48. If one corner of the rectangle is at (x, y) with x > 0, y > 0, then A = 4xy, y = 3 1 − (x/4)2 , √ √ 8 − x2 dA = 6√ A = 12x 1 − (x/4)2 = 3x 16 − x2 , , critical point at x = 2 2. Since A = 0 2 dx 16 − x √ when x = 0, 4 and A > 0 otherwise, there is an absolute maximum A = 24 at x = 2 2. (b) minimum: (−2.111985, −0.355116) maximum: (0.372591, 2.012931)

-10

2.1

10 -0.5 y 0.2 0.6

x 1

50. (a) -0.5

-1 -1.5 -2

(b) The distance between the boat and the origin is x2 + y 2 , where y = (x10/3 − 1)/(2x2/3 ). The minimum distance is 0.8247 mi when x = 0.6598 mi. The boat gets swept downstream. (c) Use the equation of the path to obtain dy/dt = (dy/dx)(dx/dt), dx/dt = (dy/dt)/(dy/dx). Let dy/dt = −4 and ﬁnd the value of dy/dx for the value of x obtained in part (b) to get dx/dt = −3 mi/h.

51. Solve φ − 0.0167 sin φ = 2π(90)/365 to get φ = 1.565978 so r = 150 × 106 (1 − 0.0167 cos φ) = 149.988 × 106 km.

uh a

52. Solve φ − 0.0934 sin φ = 2π(1)/1.88 to get φ = 3.325078 so r = 228 × 106 (1 − 0.0934 cos φ) = 248.938 × 106 km.

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Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH CHAPTER 6

Integration

us uf

EXERCISE SET 6.1 1 2 n−1 1. Endpoints 0, , , . . . , , 1; using right endpoints, n n n 1 1 2 n−1 + ··· + +1 + An = n n n n 5 0.749739

10 50 0.710509 0.676095

100 0.671463

n−1 1 2 2. Endpoints 0, , , . . . , , 1; using right endpoints, n n n n n n 1 1 n + + + ··· + + An = n+1 n+2 n+3 2n − 1 2 n 2 0.583333

3. Endpoints 0,

5 0.645635

10 50 0.668771 0.688172

100 0.690653

n An

π 2π (n − 1)π , ,..., , π; using right endpoints, n n n

100 1.99984

(n − 1)π π π 2π , ,..., , ; using right endpoints, 2n 2 2n 2n

4. Endpoints 0,

10 50 1.98352 1.99935

π n

2 5 1.57080 1.93376

An = [sin(π/n) + sin(2π/n) + · · · + sin(π(n − 1)/n) + sin π] n An

2 0.853553

n An

An = [cos(π/2n) + cos(2π/2n) + · · · + cos((n − 1)π/2n) + cos(π/2)] 2 0.555359

5 0.834683

10 50 0.919405 0.984204

100 0.992120

n An

π 2n

2 0.583333

uh a

n An

n+1 n+2 2n − 1 , 5. Endpoints 1, ,..., , 2; using right endpoints, n n n n n n 1 1 + + ··· + + An = n+1 n+2 2n − 1 2 n 5 0.645635

10 50 0.668771 0.688172

100 0.690653

π π π π 2π π (n − 1)π π 6. Endpoints − , − + , − + ,...,− + , ; using right endpoints, 2 2 n 2 n 2 n 2 π π π π π 2π π (n − 1)π + cos − + + · · · + cos − + + cos An = cos − + 2 n 2 n 2 n 2 n n An

2 5 1.99985 1.93376

10 50 1.98352 1.99936

100 1.99985

223

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224

Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Chapter 6

2 0.433013

5 0.659262

10 0.726130

50 0.774567

us uf

n An

nâ&#x2C6;&#x2019;1 1 2 , 1; using right endpoints, 7. Endpoints 0, , , . . . , n n n ďŁŽ

ďŁš

2 2

2 1 2 nâ&#x2C6;&#x2019;1 1 An = ďŁ° 1 â&#x2C6;&#x2019; + 1â&#x2C6;&#x2019; + ÂˇÂˇÂˇ + 1 â&#x2C6;&#x2019; + 0ďŁť n n n n 100 0.780106

2 1

5 1.423837

9. 3(x â&#x2C6;&#x2019; 1)

10 1.518524

50 100 1.566097 1.569136

10. 5(x â&#x2C6;&#x2019; 2)

11. x(x + 2)

13. (x + 3)(x â&#x2C6;&#x2019; 1)

14.

12.

n An

4 2(n â&#x2C6;&#x2019; 1) 2 , 1; using right endpoints, 8. Endpoints â&#x2C6;&#x2019;1, â&#x2C6;&#x2019;1 + , â&#x2C6;&#x2019;1 + , . . . , â&#x2C6;&#x2019;1 + n n n ďŁš ďŁŽ

2 2

2 nâ&#x2C6;&#x2019;4 nâ&#x2C6;&#x2019;2 2 nâ&#x2C6;&#x2019;2 An = ďŁ° 1 â&#x2C6;&#x2019; + ÂˇÂˇÂˇ + 1 â&#x2C6;&#x2019; + 0ďŁť + 1â&#x2C6;&#x2019; n n n n

3 (x â&#x2C6;&#x2019; 1)2 2

3 x(x â&#x2C6;&#x2019; 2) 2

15. The area in Exercise 13 is always 3 less than the area in Exercise 11. The regions are identical except that the area in Exercise 11 has the extra trapezoid with vertices at (0, 0), (1, 0), (0, 2), (1, 4) (with area 3). 16. (a) The region in question is a trapezoid, and the area of a trapezoid is

1 (h1 + h2 )w. 2

1 1 [f (a) + f (x)] + (x â&#x2C6;&#x2019; a) f (x) 2 2 1 f (x) â&#x2C6;&#x2019; f (a) 1 = f (x) = [f (a) + f (x)] + (x â&#x2C6;&#x2019; a) 2 2 xâ&#x2C6;&#x2019;a â&#x2C6;&#x161; 17. B is also the area between the graph of f (x) = x and the interval [0, 1] on the yâ&#x2C6;&#x2019;axis, so A + B is the area of the square.

(b) From Part (a), A (x) =

18. If the plane is rotated about the line y = x then A becomes B and vice versa.

EXERCISE SET 6.2

x dx = 1 + x2 + C 2 1+x

1. (a)

â&#x2C6;&#x161;

(b)

(x + 1)ex dx = xex + C

uh a

d (sin x â&#x2C6;&#x2019; x cos x + C) = cos x â&#x2C6;&#x2019; cos x + x sin x = x sin x dx â&#x2C6;&#x161; â&#x2C6;&#x161; d x 1 â&#x2C6;&#x2019; x2 + x2 / 1 â&#x2C6;&#x2019; x2 1 â&#x2C6;&#x161; (b) +C = = 2 2 dx 1 â&#x2C6;&#x2019; x (1 â&#x2C6;&#x2019; x2 )3/2 1â&#x2C6;&#x2019;x

2. (a)

3x2 d 3 x +5 = â&#x2C6;&#x161; dx 2 x3 + 5

3x2 â&#x2C6;&#x161; dx = x3 + 5 + C 2 x3 + 5

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Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Exercise Set 6.2

d [sin x â&#x2C6;&#x2019; x cos x] = x sin x dx

x sin x dx = sin x â&#x2C6;&#x2019; x cos x + C

(b)

3 5/3 x +C 5 1 1 xâ&#x2C6;&#x2019;3 dx = â&#x2C6;&#x2019; xâ&#x2C6;&#x2019;2 + C 2 4

7 12/7 x +C 12

(c)

1 1 (b) â&#x2C6;&#x2019; xâ&#x2C6;&#x2019;5 + C = â&#x2C6;&#x2019; 5 + C 5x 5

2 9/2 x +C 9

(b) u4 /4 â&#x2C6;&#x2019; u2 + 7u + C

9. (a)

â&#x2C6;&#x161; â&#x2C6;&#x161; cos (2 x) â&#x2C6;&#x161; dx = sin 2 x + C x

7. (a) x9 /9 + C 8. (a)

3 â&#x2C6;&#x2019; x2 x dx = 2 +C 2 2 (x + 3) x +3

â&#x2C6;&#x161; d â&#x2C6;&#x161; cos (2 x) â&#x2C6;&#x161; sin 2 x = dx x

us uf

x 3 â&#x2C6;&#x2019; x2 d = 2 2 dx x + 3 (x + 3)2

3 5/3 x â&#x2C6;&#x2019; 5x4/5 + 4x + C 5 12 1 1 2 11. (xâ&#x2C6;&#x2019;3 + x1/2 â&#x2C6;&#x2019; 3x1/4 + x2 )dx = â&#x2C6;&#x2019; xâ&#x2C6;&#x2019;2 + x3/2 â&#x2C6;&#x2019; x5/4 + x3 + C 3 5 3 2 3 8 (7y â&#x2C6;&#x2019;3/4 â&#x2C6;&#x2019; y 1/3 + 4y 1/2 )dy = 28y 1/4 â&#x2C6;&#x2019; y 4/3 + y 3/2 + C 4 3

10.

12.

4 1 (4 + 4y 2 + y 4 )dy = 4y + y 3 + y 5 + C 3 5

17.

1 2 1 (2 â&#x2C6;&#x2019; x + 2x2 â&#x2C6;&#x2019; x3 )dx = 2x â&#x2C6;&#x2019; x2 + x3 â&#x2C6;&#x2019; x4 + C 2 3 4 (x + 2xâ&#x2C6;&#x2019;2 â&#x2C6;&#x2019; xâ&#x2C6;&#x2019;4 )dx = x2 /2 â&#x2C6;&#x2019; 2/x + 1/(3x3 ) + C 1 (tâ&#x2C6;&#x2019;3 â&#x2C6;&#x2019; 2)dt = â&#x2C6;&#x2019; tâ&#x2C6;&#x2019;2 â&#x2C6;&#x2019; 2t + C 2 2 x + 3e dx = 2 ln |x| + 3ex + C x

â&#x2C6;&#x161; 1 â&#x2C6;&#x2019;1 â&#x2C6;&#x161; t 1 t â&#x2C6;&#x2019; 2e dt = ln |t| â&#x2C6;&#x2019; 2et + C 2 2

uh a

19.

12 7/3 3 x + x10/3 + C 7 10

(4x1/3 â&#x2C6;&#x2019; 4x4/3 + x7/3 )dx = 3x4/3 â&#x2C6;&#x2019;

16.

18.

x1/3 (4 â&#x2C6;&#x2019; 4x + x2 )dx =

15.

14.

(x + x4 )dx = x2 /2 + x5 /5 + C

13.

20.

225

21. â&#x2C6;&#x2019;4 cos x + 2 sin x + C 22. 4 tan x â&#x2C6;&#x2019; csc x + C 23. (sec2 x + sec x tan x)dx = tan x + sec x + C

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226

Chapter 6

24.

(sec x tan x + 1)dx = sec x + x + C

25.

sin y dy = â&#x2C6;&#x2019; cos y + C

27.

(Ď&#x2020; + 2 csc2 Ď&#x2020;)dĎ&#x2020; = Ď&#x2020;2 /2 â&#x2C6;&#x2019; 2 cot Ď&#x2020; + C

29.

sec2 Î¸ dÎ¸ = tan Î¸ + C

sec x tan x dx = sec x + C

2 sin x cos x dx = 2 cos x

30.

sin x dx = â&#x2C6;&#x2019;2 cos x + C

1 3 â&#x2C6;&#x2019; 2 1 + x2 2 1â&#x2C6;&#x2019;x

â&#x2C6;&#x161;

31.

(1 + sin Î¸)dÎ¸ = Î¸ â&#x2C6;&#x2019; cos Î¸ + C

dx =

28.

26.

us uf

sec Î¸ dÎ¸ = cos Î¸

1 sinâ&#x2C6;&#x2019;1 x â&#x2C6;&#x2019; 3 tanâ&#x2C6;&#x2019;1 x + C 2

4 1 1 1 + x + x3 â&#x2C6;&#x2019;1 â&#x2C6;&#x161; dx = 4 secâ&#x2C6;&#x2019;1 x+ x2 +tanâ&#x2C6;&#x2019;1 x+C dx = 4 sec x+ x + + 1 + x2 x2 + 1 2 x x2 â&#x2C6;&#x2019; 1 2 1 â&#x2C6;&#x2019; sin x 1 â&#x2C6;&#x2019; sin x 33. dx = sec x â&#x2C6;&#x2019; sec x tan x dx = tan x â&#x2C6;&#x2019; sec x + C dx = 2 2 cos x 1 â&#x2C6;&#x2019; sin x 1 1 1 1 sec2 x dx = tan x + C 34. dx = dx = 1 + cos 2x 2 cos2 x 2 2

32.

(b)

x 2

â&#x20AC;&#x201C;2

-1

36. (a)

35. (a)

(b)

x 1

37.

uh a

c/4

c/2

x 1

38. 2

x 1

â&#x20AC;&#x201C;4

-5

39. f (x) = m = â&#x2C6;&#x2019; sin x so f (x) =

(â&#x2C6;&#x2019; sin x)dx = cos x + C; f (0) = 2 = 1 + C

so C = 1, f (x) = cos x + 1

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Exercise Set 6.2

227

1 (x + 1)3 + C; 3 1 25 1 1 25 1 , f (x) = (x + 1)3 + f (â&#x2C6;&#x2019;2) = 8 = (â&#x2C6;&#x2019;2 + 1)3 + C = â&#x2C6;&#x2019; + C, = 8 + = 3 3 3 3 3 3 5 3 3 3 5 41. (a) y(x) = x1/3 dx = x4/3 + C, y(1) = + C = 2, C = ; y(x) = x4/3 + 4 4 4 4 4 Ï&#x20AC; 1 Ï&#x20AC; Ï&#x20AC; = â&#x2C6;&#x2019; + + C = 1/2, C = 1 â&#x2C6;&#x2019; ; (b) y(t) = (sin t + 1) dt = â&#x2C6;&#x2019; cos t + t + C, y 3 2 3 3 Ï&#x20AC; y(t) = â&#x2C6;&#x2019; cos t + t + 1 â&#x2C6;&#x2019; 3 8 8 2 (c) y(x) = (x1/2 + xâ&#x2C6;&#x2019;1/2 )dx = x3/2 + 2x1/2 + C, y(1) = 0 = + C, C = â&#x2C6;&#x2019; , 3 3 3 2 3/2 8 y(x) = x + 2x1/2 â&#x2C6;&#x2019; 3 3 1 â&#x2C6;&#x2019;3 1 1 1 1 1 42. (a) y(x) = dx = â&#x2C6;&#x2019; xâ&#x2C6;&#x2019;2 + C, y(1) = 0 = â&#x2C6;&#x2019; + C, C = x ; y(x) = â&#x2C6;&#x2019; xâ&#x2C6;&#x2019;2 + 8 16 16 16 16 16 â&#x2C6;&#x161; â&#x2C6;&#x161; 2 2 Ï&#x20AC; (b) y(t) = (sec2 t â&#x2C6;&#x2019; sin t) dt = tan t + cos t + C, y( ) = 1 = 1 + + C, C = â&#x2C6;&#x2019; ; 4 2 2 â&#x2C6;&#x161; 2 y(t) = tan t + cos t â&#x2C6;&#x2019; 2 2 2 (c) y(x) = x7/2 dx = x9/2 + C, y(0) = 0 = C, C = 0; y(x) = x9/2 9 9 43. (a) y = 4ex dx = 4ex + C, 1 = y(0) = 4 + C, C = â&#x2C6;&#x2019;3, y = 4ex â&#x2C6;&#x2019; 3 (b) y(t) = tâ&#x2C6;&#x2019;1 dt = ln |t| + C, y(â&#x2C6;&#x2019;1) = C = 5, C = 5; y(t) = ln |t| + 5

(x + 1)2 dx =

us uf

40. f (x) = m = (x + 1) , so f (x) =

â&#x2C6;&#x161; 3 3 â&#x2C6;&#x2019;1 â&#x2C6;&#x161; 44. (a) y = = 0 = Ï&#x20AC; + C, C = â&#x2C6;&#x2019;Ï&#x20AC;, y = 3 sinâ&#x2C6;&#x2019;1 t â&#x2C6;&#x2019; Ï&#x20AC; dt = 3 sin t + C, y 2 2 1â&#x2C6;&#x2019;t dy 2 2 dx = x â&#x2C6;&#x2019; 2 tanâ&#x2C6;&#x2019;1 x + C, (b) =1â&#x2C6;&#x2019; 2 ,y = 1â&#x2C6;&#x2019; 2 dx x +1 x +1 Ï&#x20AC; Ï&#x20AC; y(1) = = 1 â&#x2C6;&#x2019; 2 + C, C = Ï&#x20AC; â&#x2C6;&#x2019; 1, y = x â&#x2C6;&#x2019; 2 tanâ&#x2C6;&#x2019;1 x + Ï&#x20AC; â&#x2C6;&#x2019; 1 2 4

2 3/2 4 5/2 x + C1 ; f (x) = x + C1 x + C2 3 15

45. f (x) =

46. f (x) = x2 /2 + sin x + C1 , use f (0) = 2 to get C1 = 2 so f (x) = x2 /2 + sin x + 2, f (x) = x3 /6 â&#x2C6;&#x2019; cos x + 2x + C2 , use f (0) = 1 to get C2 = 2 so f (x) = x3 /6 â&#x2C6;&#x2019; cos x + 2x + 2

uh a

47. dy/dx = 2x + 1, y =

(2x + 1)dx = x2 + x + C; y = 0 when x = â&#x2C6;&#x2019;3

so (â&#x2C6;&#x2019;3)2 + (â&#x2C6;&#x2019;3) + C = 0, C = â&#x2C6;&#x2019;6 thus y = x2 + x â&#x2C6;&#x2019; 6

48. dy/dx = x2 , y =

x2 dx = x3 /3 + C; y = 2 when x = â&#x2C6;&#x2019;1 so (â&#x2C6;&#x2019;1)3 /3 + C = 2, C = 7/3

thus y = x3 /3 + 7/3

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228

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Chapter 6

6xdx = 3x2 + C1 . The slope of the tangent line is â&#x2C6;&#x2019;3 so dy/dx = â&#x2C6;&#x2019;3 when x = 1. 2 2 Thus 3(1) + C1 = â&#x2C6;&#x2019;3, C1 = â&#x2C6;&#x2019;6 so dy/dx = 3x â&#x2C6;&#x2019; 6, y = (3x2 â&#x2C6;&#x2019; 6)dx = x3 â&#x2C6;&#x2019; 6x + C2 . If x = 1,

us uf

then y = 5 â&#x2C6;&#x2019; 3(1) = 2 so (1)2 â&#x2C6;&#x2019; 6(1) + C2 = 2, C2 = 7 thus y = x3 â&#x2C6;&#x2019; 6x + 7.

49. dy/dx =

51. (a) F (x) = G (x) = 3x + 4 (b) F (0) = 16/6 = 8/3, G(0) = 0, so F (0) â&#x2C6;&#x2019; G(0) = 8/3 (c) F (x) = (9x2 + 24x + 16)/6 = 3x2 /2 + 4x + 8/3 = G(x) + 8/3

50. dT /dx = C1 , T = C1 x + C2 ; T = 25 when x = 0 so C2 = 25, T = C1 x + 25. T = 85 when x = 50 so 50C1 + 25 = 85, C1 = 1.2, T = 1.2x + 25

1 (1 â&#x2C6;&#x2019; cos x)dx = (x â&#x2C6;&#x2019; sin x) + C 2

(b)

(b) G(x) â&#x2C6;&#x2019; F (x) =

2, 3,

1 2

(1 + cos x) dx =

1 (x + sin x) + C 2

1, x > 0 â&#x2C6;&#x2019;1, x < 0

56. (a) F (x) = G (x) = f (x), where f (x) =

55. (a)

1 2

52. (a) F (x) = G (x) = 10x/(x2 + 5)2 (b) F (0) = 0, G(0) = â&#x2C6;&#x2019;1, so F (0) â&#x2C6;&#x2019; G(0) = 1 (x2 + 5) â&#x2C6;&#x2019; 5 5 x2 = =1â&#x2C6;&#x2019; 2 (c) F (x) = 2 = G(x) + 1 x +5 x +5 x2 + 5 54. (csc2 x â&#x2C6;&#x2019; 1)dx = â&#x2C6;&#x2019; cot x â&#x2C6;&#x2019; x + C 53. (sec2 x â&#x2C6;&#x2019; 1)dx = tan x â&#x2C6;&#x2019; x + C

x>0 so G(x) = F (x) plus a constant x<0

EXERCISE SET 6.3

(c) no, because (â&#x2C6;&#x2019;â&#x2C6;&#x17E;, 0) â&#x2C6;Ş (0, +â&#x2C6;&#x17E;) is not an interval 1087 1/2 1087 1/2 1087 ft/s T + C, v(273) = 1087 = 1087 + C so C = 0, v = â&#x2C6;&#x161; T T â&#x2C6;&#x2019;1/2 dT = â&#x2C6;&#x161; 57. v = â&#x2C6;&#x161; 273 273 2 273

u23 du = u24 /24 + C = (x2 + 1)24 /24 + C (b) â&#x2C6;&#x2019; u3 du = â&#x2C6;&#x2019;u4 /4 + C = â&#x2C6;&#x2019;(cos4 x)/4 + C â&#x2C6;&#x161; (c) 2 sin u du = â&#x2C6;&#x2019;2 cos u + C = â&#x2C6;&#x2019;2 cos x + C 3 3 2 3 uâ&#x2C6;&#x2019;1/2 du = u1/2 + C = 4x + 5 + C (d) 8 4 4

1. (a)

1 1 1 sec2 u du = tan u + C = tan(4x + 1) + C 4 4 4 1 1 1 u1/2 du = u3/2 + C = (1 + 2y 2 )3/2 + C (b) 4 6 6 2 3/2 2 1 u1/2 du = (c) u +C = sin3/2 (Ď&#x20AC;Î¸) + C Ď&#x20AC; 3Ď&#x20AC; 3Ď&#x20AC; 5 5 (d) u4/5 du = u9/5 + C = (x2 + 7x + 3)9/5 + C 9 9

uh a

2. (a)

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Exercise Set 6.3

(u â&#x2C6;&#x2019; 1) u

4. (a)

us uf

2 1/2

(u5/2 â&#x2C6;&#x2019; 2u3/2 + u1/2 )du =

du = =

2 7/2 4 5/2 2 3/2 u â&#x2C6;&#x2019; u + u +C 7 5 3

2 4 2 (1 + x)7/2 â&#x2C6;&#x2019; (1 + x)5/2 + (1 + x)3/2 + C 7 5 3

1 1 3. (a) â&#x2C6;&#x2019; u du = â&#x2C6;&#x2019; u2 + C = â&#x2C6;&#x2019; cot2 x + C 2 2 1 1 (b) u9 du = u10 + C = (1 + sin t)10 + C 10 10 1 1 1 cos u du = sin u + C = sin 2x + C (c) 2 2 2 1 1 1 sec2 u du = tan u + C = tan x2 + C (d) 2 2 2

csc2 u du = â&#x2C6;&#x2019; cot u + C = â&#x2C6;&#x2019; cot(sin x) + C

(b)

sin u du = â&#x2C6;&#x2019; cos u + C = â&#x2C6;&#x2019; cos(x â&#x2C6;&#x2019; Ď&#x20AC;) + C

(d)

du 1 1 =â&#x2C6;&#x2019; +C =â&#x2C6;&#x2019; 5 +C u2 u x +1

(c)

1 du = ln |u| + C = ln | ln x| + C u 1 1 1 eu du = â&#x2C6;&#x2019; eu + C = â&#x2C6;&#x2019; eâ&#x2C6;&#x2019;5x + C (b) â&#x2C6;&#x2019; 5 5 5 1 1 1 1 du = â&#x2C6;&#x2019; ln |u| + C = â&#x2C6;&#x2019; ln |1 + cos 3Î¸| + C (c) â&#x2C6;&#x2019; 3 u 3 3 du (d) = ln u + C = ln(1 + ex ) + C u

5. (a)

du 1 1 = tanâ&#x2C6;&#x2019;1 (x3 ) + C 3 1 + u2 3 1 â&#x2C6;&#x161; (b) u = ln x, du = sinâ&#x2C6;&#x2019;1 (ln x) + C 2 1 â&#x2C6;&#x2019; u 1 â&#x2C6;&#x161; du = secâ&#x2C6;&#x2019;1 (3x) + C (c) u = 3x, 2â&#x2C6;&#x2019;1 u u â&#x2C6;&#x161; â&#x2C6;&#x161; du (d) u = x, 2 = 2 tanâ&#x2C6;&#x2019;1 u + C = 2 tanâ&#x2C6;&#x2019;1 ( x) + C 1 + u2

6. (a) u = x3 ,

1 7. u = 2 â&#x2C6;&#x2019; x , du = â&#x2C6;&#x2019;2x dx; â&#x2C6;&#x2019; 2

uh a

1 8. u = 3x â&#x2C6;&#x2019; 1, du = 3dx; 3 1 9. u = 8x, du = 8dx; 8

1 10. u = 3x, du = 3dx; 3

u3 du = â&#x2C6;&#x2019;u4 /8 + C = â&#x2C6;&#x2019;(2 â&#x2C6;&#x2019; x2 )4 /8 + C

u5 du =

cos u du =

229

1 6 1 u +C = (3x â&#x2C6;&#x2019; 1)6 + C 18 18

1 1 sin u + C = sin 8x + C 8 8

1 1 sin u du = â&#x2C6;&#x2019; cos u + C = â&#x2C6;&#x2019; cos 3x + C 3 3

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Chapter 6

12. u = 5x, du = 5dx;

1 5

13. u = 2x, du = 2dx;

1 2

19.

20.

21.

22.

23.

eu du =

1 u 1 e + C = e2x + C 2 2

1 1 1 1 du = ln |u| + C = ln |2x| + C u = 2x, du = 2dx; 2 u 2 2 1 1 1 â&#x2C6;&#x161; u = 2x, du = sinâ&#x2C6;&#x2019;1 (2x) + C 2 2 1 â&#x2C6;&#x2019; u2 1 1 1 du = tanâ&#x2C6;&#x2019;1 (4x) + C u = 4x, 4 1 + u2 4 1 1 3/2 1 2 (7t2 + 12)3/2 + C u1/2 du = u +C = u = 7t + 12, du = 14t dt; 21 14 21 1 1 1 uâ&#x2C6;&#x2019;1/2 du = â&#x2C6;&#x2019; u1/2 + C = â&#x2C6;&#x2019; u = 4 â&#x2C6;&#x2019; 5x2 , du = â&#x2C6;&#x2019;10x dx; â&#x2C6;&#x2019; 4 â&#x2C6;&#x2019; 5x2 + C 5 5 10 1 2 2 3 uâ&#x2C6;&#x2019;1/2 du = u1/2 + C = u = x3 + 1, du = 3x2 dx; x +1+C 3 3 3 1 1 1 uâ&#x2C6;&#x2019;2 du = uâ&#x2C6;&#x2019;1 + C = (1 â&#x2C6;&#x2019; 3x)â&#x2C6;&#x2019;1 + C u = 1 â&#x2C6;&#x2019; 3x, du = â&#x2C6;&#x2019;3dx; â&#x2C6;&#x2019; 3 3 3 1 1 1 uâ&#x2C6;&#x2019;3 du = â&#x2C6;&#x2019; uâ&#x2C6;&#x2019;2 + C = â&#x2C6;&#x2019; (4x2 + 1)â&#x2C6;&#x2019;2 + C u = 4x2 + 1, du = 8x dx; 8 16 16 1 1 1 cos u du = sin u + C = sin(3x2 ) + C u = 3x2 , du = 6x dx; 6 6 6 u = sin x, du = cos x dx; eu du = eu + C = esin x + C

18.

1 1 tan u + C = tan 5x + C 5 5

17.

sec2 u du =

16.

1 1 sec u + C = sec 4x + C 4 4

15.

sec u tan u du =

1 4

14.

1 11. u = 4x, du = 4dx; 4

us uf

230

24. u = x4 , du = 4x3 dx;

eu du =

1 25. u = â&#x2C6;&#x2019;2x , du = â&#x2C6;&#x2019;6x , â&#x2C6;&#x2019; 6

1 u 1 4 e + C = ex + C 4 4

3 1 1 eu du = â&#x2C6;&#x2019; eu + C = â&#x2C6;&#x2019; eâ&#x2C6;&#x2019;2x + C 6 6 1 du = ln |u| + C = ln ex â&#x2C6;&#x2019; eâ&#x2C6;&#x2019;x + C 26. u = ex â&#x2C6;&#x2019; eâ&#x2C6;&#x2019;x , du = (ex + eâ&#x2C6;&#x2019;x )dx, u 1 1 1 â&#x2C6;&#x2019;1 x 2 1 27. u = ex , du = tan (e ) + C 28. u = t , du = tanâ&#x2C6;&#x2019;1 (t2 ) + C 2 2 1+u 2 u +1 2

uh a

1 1 1 sin u du = cos u + C = cos(5/x) + C 29. u = 5/x, du = â&#x2C6;&#x2019;(5/x )dx; â&#x2C6;&#x2019; 5 5 5 â&#x2C6;&#x161; â&#x2C6;&#x161; 1 30. u = x, du = â&#x2C6;&#x161; dx; 2 sec2 u du = 2 tan u + C = 2 tan x + C 2 x

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Exercise Set 6.3

u5 du =

1 34. u = 5 + cos 2Î¸, du = â&#x2C6;&#x2019;2 sin 2Î¸ dÎ¸; â&#x2C6;&#x2019; 2 1 35. u = 2 â&#x2C6;&#x2019; sin 4Î¸, du = â&#x2C6;&#x2019;4 cos 4Î¸ dÎ¸; â&#x2C6;&#x2019; 4 1 36. u = tan 5x, du = 5 sec 5x dx; 5

38. u = cos Î¸, â&#x2C6;&#x2019;

1 du = â&#x2C6;&#x2019; tanâ&#x2C6;&#x2019;1 (cos Î¸) + C u2 + 1 u2 du =

dx; u = â&#x2C6;&#x2019;x, du = â&#x2C6;&#x2019;dx; â&#x2C6;&#x2019;

x/2

43. u =

eu du = â&#x2C6;&#x2019;eu + C = â&#x2C6;&#x2019;eâ&#x2C6;&#x2019;x + C

â&#x2C6;&#x161;

1 dy, 2 y + 1, du = â&#x2C6;&#x161; 2 y+1

â&#x2C6;&#x161;

1 y, du = â&#x2C6;&#x161; dy, 2 2 y

â&#x2C6;&#x161; eu du = 2eu + C = 2ex/2 + C = 2 ex + C

â&#x2C6;&#x161; y+1

eu du = 2eu + C = 2e

1 du = 2 eu

+C â&#x2C6;&#x161; y

eâ&#x2C6;&#x2019;u du = â&#x2C6;&#x2019;2eâ&#x2C6;&#x2019;u + C = â&#x2C6;&#x2019;2eâ&#x2C6;&#x2019;

44. u =

dx; u = x/2, du = dx/2; 2

42.

1 3 1 u + C = sec3 2x + C 6 6

sin u du = â&#x2C6;&#x2019; cos u + C = â&#x2C6;&#x2019; cos(sin Î¸) + C

40. u = sin Î¸, du = cos Î¸ dÎ¸;

41.

1 4 1 u +C = tan4 5x + C 20 20

1 du = sinâ&#x2C6;&#x2019;1 (tan x) + C 1 â&#x2C6;&#x2019; u2

â&#x2C6;&#x2019;x

1 1 u1/2 du = â&#x2C6;&#x2019; u3/2 + C = â&#x2C6;&#x2019; (2 â&#x2C6;&#x2019; sin 4Î¸)3/2 + C 6 6

u3 du =

1 39. u = sec 2x, du = 2 sec 2x tan 2x dx; 2

1 â&#x2C6;&#x2019;2 1 u + C = (5 + cos 2Î¸)â&#x2C6;&#x2019;2 + C 4 4

uâ&#x2C6;&#x2019;3 du =

â&#x2C6;&#x161;

1 6 1 u +C = sin6 3t + C 18 18

37. u = tan x,

1 1 u3 du = â&#x2C6;&#x2019; u4 + C = â&#x2C6;&#x2019; cos4 2t + C 8 8

us uf

1 33. u = sin 3t, du = 3 cos 3t dt; 3

1 32. u = cos 2t, du = â&#x2C6;&#x2019;2 sin 2t dt; â&#x2C6;&#x2019; 2

1 1 tan u + C = tan(x3 ) + C 3 3

sec2 u du =

1 31. u = x , du = 3x dx; 3 3

45. u = x â&#x2C6;&#x2019; 3, x = u + 3, dx = du 2 2 (u + 3)u1/2 du = (u3/2 + 3u1/2 )du = u5/2 + 2u3/2 + C = (x â&#x2C6;&#x2019; 3)5/2 + 2(x â&#x2C6;&#x2019; 3)3/2 + C 5 5

uh a

46. u = y + 1, y = u â&#x2C6;&#x2019; 1, dy = du uâ&#x2C6;&#x2019;1 2 2 du = (u1/2 â&#x2C6;&#x2019; uâ&#x2C6;&#x2019;1/2 )du = u3/2 â&#x2C6;&#x2019; 2u1/2 + C = (y + 1)3/2 â&#x2C6;&#x2019; 2(y + 1)1/2 + C 1/2 3 3 u

47.

231

sin 2Î¸ sin 2Î¸ dÎ¸ = (1 â&#x2C6;&#x2019; cos2 2Î¸) sin 2Î¸ dÎ¸; u = cos 2Î¸, du = â&#x2C6;&#x2019;2 sin 2Î¸ dÎ¸, 1 1 1 1 1 (1 â&#x2C6;&#x2019; u2 )du = â&#x2C6;&#x2019; u + u3 + C = â&#x2C6;&#x2019; cos 2Î¸ + cos3 2Î¸ + C â&#x2C6;&#x2019; 2 2 6 2 6

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Chapter 6

52.

cos x dx; u = sin x, du = cos xdx; sin x

53. (a) sinâ&#x2C6;&#x2019;1 â&#x2C6;&#x161; (x/3) + C â&#x2C6;&#x161; (c) (1/ Ď&#x20AC;) secâ&#x2C6;&#x2019;1 (x/ Ď&#x20AC;) + C

1 du = ln |u| + C = ln | sin x| + C u â&#x2C6;&#x161; â&#x2C6;&#x161; (b) (1/ 5) tanâ&#x2C6;&#x2019;1 (x/ 5) + C

1 1 du = tanâ&#x2C6;&#x2019;1 (ex /2) + C 2 4 + u2 1 1 1 â&#x2C6;&#x161; du = sinâ&#x2C6;&#x2019;1 (2x/3) + C, (b) u = 2x, 2 2 2 9â&#x2C6;&#x2019;u â&#x2C6;&#x161; â&#x2C6;&#x161; â&#x2C6;&#x161; 1 1 â&#x2C6;&#x161; (c) u = 5y, du = â&#x2C6;&#x161; secâ&#x2C6;&#x2019;1 ( 5y/ 3) + C 2 3 u u â&#x2C6;&#x2019;3

54. (a) u = ex ,

us uf

48. sec2 3Î¸ = tan2 3Î¸ + 1, u = 3Î¸, du = 3dÎ¸ 1 1 1 1 1 sec4 3Î¸ dÎ¸ = (tan2 u + 1) sec2 u du = tan3 u + tan u + C = tan3 3Î¸ + tan 3Î¸ + C 3 9 3 9 3 1 49. 1+ dt = t + ln |t| + C t 1 2 ln x 2 ln x ln x2 2 50. e =e = x , x > 0, so e dx = x2 dx = x3 + C 3 51. ln(ex ) + ln(eâ&#x2C6;&#x2019;x ) = ln(ex eâ&#x2C6;&#x2019;x ) = ln 1 = 0 so [ln(ex ) + ln(eâ&#x2C6;&#x2019;x )]dx = C

232

55. u = a + bx, du = bdx, 1 (a + bx)n+1 n (a + bx) dx = un du = +C b b(n + 1)

1 56. u = a + bx, du = b dx, dx = du b 1 n n 1/n (n+1)/n u du = +C = u (a + bx)(n+1)/n + C b b(n + 1) b(n + 1)

57. u = sin(a + bx), du = b cos(a + bx)dx 1 1 1 un du = un+1 + C = sinn+1 (a + bx) + C b(n + 1) b(n + 1) b 1 1 59. (a) with u = sin x, du = cos x dx; u du = u2 + C1 = sin2 x + C1 ; 2 2 1 1 with u = cos x, du = â&#x2C6;&#x2019; sin x dx; â&#x2C6;&#x2019; u du = â&#x2C6;&#x2019; u2 + C2 = â&#x2C6;&#x2019; cos2 x + C2 2 2

uh a

(b) because they diďŹ&#x20AC;er by a constant: 1 1 1 2 2 sin x + C1 â&#x2C6;&#x2019; â&#x2C6;&#x2019; cos x + C2 = (sin2 x + cos2 x) + C1 â&#x2C6;&#x2019; C2 = 1/2 + C1 â&#x2C6;&#x2019; C2 2 2 2 25 3 60. (a) First method: (25x2 â&#x2C6;&#x2019; 10x + 1)dx = x â&#x2C6;&#x2019; 5x2 + x + C1 ; 3 1 1 3 1 second method: u2 du = u + C2 = (5x â&#x2C6;&#x2019; 1)3 + C2 5 15 15

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Exercise Set 6.3

y(1) =

â&#x2C6;&#x161;

3x + 1dx =

2 (3x + 1)3/2 + C, 9

29 2 29 16 + C = 5, C = so y(x) = (3x + 1)3/2 + 9 9 9 9 (6 â&#x2C6;&#x2019; 5 sin 2x)dx = 6x +

62. y(x) = y(0) =

us uf

61. y(x) =

1 1 25 3 1 (5x â&#x2C6;&#x2019; 1)3 + C2 = (125x3 â&#x2C6;&#x2019; 75x2 + 15x â&#x2C6;&#x2019; 1) + C2 = x â&#x2C6;&#x2019; 5x2 + x â&#x2C6;&#x2019; + C2 ; 15 15 3 15 the answers diďŹ&#x20AC;er by a constant.

5 cos 2x + C, 2

(b)

233

1 5 1 5 + C = 3, C = so y(x) = 6x + cos 2x + 2 2 2 2

2 2 2eâ&#x2C6;&#x2019;t dt = â&#x2C6;&#x2019;2eâ&#x2C6;&#x2019;t + C, y(1) = â&#x2C6;&#x2019; + C = 3 â&#x2C6;&#x2019; , C = 3; y(t) = â&#x2C6;&#x2019;2eâ&#x2C6;&#x2019;t + 3 e e

63. y(t) =

dx , u = x/5, dx = 5 du, 100 + 4x2 du 3Ď&#x20AC; 1 1 1 1 Ď&#x20AC; â&#x2C6;&#x2019;1 â&#x2C6;&#x2019;1 x + C; y(â&#x2C6;&#x2019;5) = â&#x2C6;&#x2019; + C, = u + C = y= tan tan = 20 1 + u2 20 20 5 80 20 4 x 1 Ď&#x20AC; Ď&#x20AC; ,y= tanâ&#x2C6;&#x2019;1 + C= 20 20 5 20

65.

64. y =

66.

-5

â&#x2C6;&#x161;

-4

2 (3x + 1)3/2 + C 9 7 2 7 2 f (0) = 1 = + C, C = , so f (x) = (3x + 1)3/2 + 9 9 9 9

3x + 1, f (x) =

(3x + 1)1/2 dx =

67. f (x) = m =

8 8 256 (4 + 0.15t)5/2 + C; p(0) = 100,000 = 45/2 + C = + C, 3 3 3 8 8 256 â&#x2030;&#x2C6; 99,915, p(t) â&#x2030;&#x2C6; (4 + 0.15t)5/2 + 99,915, p(5) â&#x2030;&#x2C6; (4.75)5/2 + 99,915 â&#x2030;&#x2C6; 100,046 C = 100,000 â&#x2C6;&#x2019; 3 3 3 (4 + 0.15t)3/2 dt =

uh a

68. p(t) =

69. u = a sin Î¸, du = a cos Î¸ dÎ¸;

â&#x2C6;&#x161;

du u = aÎ¸ + C = sinâ&#x2C6;&#x2019;1 + C 2 a â&#x2C6;&#x2019;u

70. If u > 0 then u = a sec Î¸, du = a sec Î¸ tan Î¸ dÎ¸,

du u 1 1 = Î¸ = secâ&#x2C6;&#x2019;1 + C 2 2 a a a u u â&#x2C6;&#x2019;a â&#x2C6;&#x161;

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Chapter 6

EXERCISE SET 6.4

(b) 5 + 8 + 11 + 14 + 17 = 55 (d) 1 + 1 + 1 + 1 + 1 + 1 = 6 (f ) 0 + 0 + 0 + 0 + 0 + 0 = 0

us uf

1. (a) 1 + 8 + 27 = 36 (c) 20 + 12 + 6 + 2 + 0 + 0 = 40 (e) 1 â&#x2C6;&#x2019; 2 + 4 â&#x2C6;&#x2019; 8 + 16 = 11

(2k â&#x2C6;&#x2019; 1)

k=1

9. (a) 10. (a)

50 k=1 5

(â&#x2C6;&#x2019;1)k+1 ak

(b)

k(k 2 â&#x2C6;&#x2019; 4) =

kâ&#x2C6;&#x2019;

nâ&#x2C6;&#x2019;1

k3 =

19.

nâ&#x2C6;&#x2019;1

k=1

20.

ss k=1

k3 â&#x2C6;&#x2019; 4

k=1

20 k=1

k2 â&#x2C6;&#x2019;

k=1 3

7 (100)(101) + 100 = 35,450 2

k 2 = 2870 â&#x2C6;&#x2019; 14 = 2856

k=1

1 1 (30)2 (31)2 â&#x2C6;&#x2019; 4 Â· (30)(31) = 214,365 4 2

1 1 (6)(7) â&#x2C6;&#x2019; (6)2 (7)2 = â&#x2C6;&#x2019;420 2 4

k=1

nâ&#x2C6;&#x2019;1 k3 1 3 1 1 1 = k = 2 Â· (n â&#x2C6;&#x2019; 1)2 n2 = (n â&#x2C6;&#x2019; 1)2 n2 n2 n 4 4

2k â&#x2C6;&#x2019; n n

k=1

5 2 1 2 5 1â&#x2C6;&#x2019; k = (n) â&#x2C6;&#x2019; Â· n(n + 1) = 4 â&#x2C6;&#x2019; n n n n n 2 n

k=1

a5â&#x2C6;&#x2019;k bk

k=0 100

nâ&#x2C6;&#x2019;1 k2 1 1 1 2 1 k = Â· (n â&#x2C6;&#x2019; 1)(n)(2n â&#x2C6;&#x2019; 1) = (n â&#x2C6;&#x2019; 1)(2n â&#x2C6;&#x2019; 1) = n n n 6 6

n 5

k=1

(d)

1 k

3 1 3 3 k = Â· n(n + 1) = (n + 1) n n 2 2 n

uh a

k=1

14.

ak xk

k=0

100

k=1

(k 3 â&#x2C6;&#x2019; 4k) =

(â&#x2C6;&#x2019;1)k+1

18.

6 k=1

n 3k k=1

(2k â&#x2C6;&#x2019; 1)

k=1 n

k=1

17.

(â&#x2C6;&#x2019;1)k+1 bk

k=0

1 (20)(21)(41) = 2870 6

(b)

1 (100)(100 + 1) = 5050 2

k=1

16.

(â&#x2C6;&#x2019;1)k+1 (2k â&#x2C6;&#x2019; 1)

15.

13.

k=1

11.

k=1

2. (a) 1 + 0 â&#x2C6;&#x2019; 3 + 0 = â&#x2C6;&#x2019;2 (b) 1 â&#x2C6;&#x2019; 1 + 1 â&#x2C6;&#x2019; 1 + 1 â&#x2C6;&#x2019; 1 = 0 (c) Ï&#x20AC; 2 + Ï&#x20AC; 2 + Â· Â· Â· + Ï&#x20AC; 2 = 14Ï&#x20AC; 2 (d) 24 + 25 + 26 = 112 â&#x2C6;&#x161; (14 â&#x2C6;&#x161; terms) â&#x2C6;&#x161; â&#x2C6;&#x161; â&#x2C6;&#x161; â&#x2C6;&#x161; (e) 1+ 2+ 3+ 4+ 5+ 6 (f ) 1 â&#x2C6;&#x2019; 1 + 1 â&#x2C6;&#x2019; 1 + 1 â&#x2C6;&#x2019; 1 + 1 â&#x2C6;&#x2019; 1 + 1 â&#x2C6;&#x2019; 1 + 1 = 1

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n(n + 1) = 465, n2 + n â&#x2C6;&#x2019; 930 = 0, (n + 31)(n â&#x2C6;&#x2019; 30) = 0, n = 30. 2

23.

n n n+1 1 1 1 1 + 2 + 3 + Â·Â·Â· + n k n+1 1 = = k = 2 Â· n(n + 1) = ; lim = n2 n2 n2 n 2 2n nâ&#x2020;&#x2019;+â&#x2C6;&#x17E; 2n 2 k=1

k=1

n n 12 + 22 + 32 + Â· Â· Â· + n2 k2 1 2 1 1 (n + 1)(2n + 1) = = k = 3 Â· n(n + 1)(2n + 1) = ; 3 3 3 n n n n 6 6n2

lim

nâ&#x2020;&#x2019;+â&#x2C6;&#x17E; n 5k

26.

nâ&#x2C6;&#x2019;1 k=1

n 5(n + 1) 5(n + 1) 5 1 5 5 ; lim k = 2 Â· n(n + 1) = = 2 nâ&#x2020;&#x2019;+â&#x2C6;&#x17E; 2n 2 n n 2 2n k=1

nâ&#x2C6;&#x2019;1 2k 2 2 2 2 1 (n â&#x2C6;&#x2019; 1)(2n â&#x2C6;&#x2019; 1) ; = k = 3 Â· (n â&#x2C6;&#x2019; 1)(n)(2n â&#x2C6;&#x2019; 1) = n 6 3n2 n3 n3

lim

k=1

nâ&#x2020;&#x2019;+â&#x2C6;&#x17E;

27. (a)

2 (n â&#x2C6;&#x2019; 1)(2n â&#x2C6;&#x2019; 1) 1 (1 â&#x2C6;&#x2019; 1/n)(2 â&#x2C6;&#x2019; 1/n) = = lim nâ&#x2020;&#x2019;+â&#x2C6;&#x17E; 3 3n2 3

(b)

28. (a)

2jâ&#x2C6;&#x2019;1

(c)

j=1

(k + 4)2k+8

(b)

k=1

2jâ&#x2C6;&#x2019;2

j=2

j=0

k=1

1 (n + 1)(2n + 1) 1 (1 + 1/n)(2 + 1/n) = = lim nâ&#x2020;&#x2019;+â&#x2C6;&#x17E; 6 6n2 3

(k â&#x2C6;&#x2019; 4)2k

k=9

29. Endpoints 2, 3, 4, 5, 6; â&#x2C6;&#x2020;x = 1; 4

f (xâ&#x2C6;&#x2014;k )â&#x2C6;&#x2020;x = 7 + 10 + 13 + 16 = 46

k=1

(b) Midpoints:

(a) Left endpoints:

25.

k=1

24.

us uf

22.

f (xâ&#x2C6;&#x2014;k )â&#x2C6;&#x2020;x = 8.5 + 11.5 + 14.5 + 17.5 = 52

k=1 4

f (xâ&#x2C6;&#x2014;k )â&#x2C6;&#x2020;x = 10 + 13 + 16 + 19 = 58

k=1

uh a

30. Endpoints 1, 3, 5, 7, 9, â&#x2C6;&#x2020;x = 2; 4 352 1 1 1 â&#x2C6;&#x2014; 2= (a) Left endpoints: f (xk )â&#x2C6;&#x2020;x = 1 + + + 3 5 7 105 k=1 4 25 1 1 1 1 (b) Midpoints: + + + 2= f (xâ&#x2C6;&#x2014;k )â&#x2C6;&#x2020;x = 2 4 6 8 12 k=1 4 496 1 1 1 1 (c) Right endpoints: + + + 2= f (xâ&#x2C6;&#x2014;k )â&#x2C6;&#x2020;x = 3 5 7 9 315 k=1

31. Endpoints: 0, Ï&#x20AC;/4, Ï&#x20AC;/2, 3Ï&#x20AC;/4, Ï&#x20AC;; â&#x2C6;&#x2020;x = Ï&#x20AC;/4 4 â&#x2C6;&#x161; â&#x2C6;&#x161; (a) Left endpoints: f (xâ&#x2C6;&#x2014;k )â&#x2C6;&#x2020;x = 1 + 2/2 + 0 â&#x2C6;&#x2019; 2/2 (Ï&#x20AC;/4) = Ï&#x20AC;/4

235

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Exercise Set 6.4

k=1

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Chapter 6

(b) Midpoints:

f (xâ&#x2C6;&#x2014;k )â&#x2C6;&#x2020;x = [cos(Ï&#x20AC;/8) + cos(3Ï&#x20AC;/8) + cos(5Ï&#x20AC;/8) + cos(7Ï&#x20AC;/8)] (Ï&#x20AC;/4)

us uf

= [cos(Ï&#x20AC;/8) + cos(3Ï&#x20AC;/8) â&#x2C6;&#x2019; cos(3Ï&#x20AC;/8) â&#x2C6;&#x2019; cos(Ï&#x20AC;/8)] (Ï&#x20AC;/4) = 0 4 â&#x2C6;&#x161; â&#x2C6;&#x161; f (xâ&#x2C6;&#x2014;k )â&#x2C6;&#x2020;x = 2/2 + 0 â&#x2C6;&#x2019; 2/2 â&#x2C6;&#x2019; 1 (Ï&#x20AC;/4) = â&#x2C6;&#x2019;Ï&#x20AC;/4 (c) Right endpoints: k=1

(c)

k=1 4

5 3 3 15 =4 f (xâ&#x2C6;&#x2014;k )â&#x2C6;&#x2020;x = â&#x2C6;&#x2019; + + + 4 4 4 4 f (xâ&#x2C6;&#x2014;k )â&#x2C6;&#x2020;x = 0 + 1 + 0 â&#x2C6;&#x2019; 3 = â&#x2C6;&#x2019;2

(b)

32. Endpoints â&#x2C6;&#x2019;1, 0, 1, 2, 3; â&#x2C6;&#x2020;x = 1 4 (a) f (xâ&#x2C6;&#x2014;k )â&#x2C6;&#x2020;x = â&#x2C6;&#x2019;3 + 0 + 1 + 0 = â&#x2C6;&#x2019;2 k=1 4

k=1

33. (a) 0.718771403, 0.705803382, 0.698172179 (b) 0.668771403, 0.680803382, 0.688172179 (c) 0.692835360, 0.693069098, 0.693134682

k=1

34. (a) 0.761923639, 0.712712753, 0.684701150 (b) 0.584145862, 0.623823864, 0.649145594 (c) 0.663501867, 0.665867079, 0.666538346

35. (a) 4.884074734, 5.115572731, 5.248762738 (b) 5.684074734, 5.515572731, 5.408762738 (c) 5.34707029, 5.338362719, 5.334644416 36. (a) 0.919403170, 0.960215997, 0.984209789 (b) 1.076482803, 1.038755813, 1.015625715 (c) 1.001028824, 1.000257067, 1.000041125

3 â&#x2C6;&#x2014; 3 3 1 â&#x2C6;&#x2014; 1 3 3 1 3 â&#x2C6;&#x2014; 37. â&#x2C6;&#x2020;x = , xk = 1 + k; f (xk )â&#x2C6;&#x2020;x = xk â&#x2C6;&#x2020;x = 1+ k k = + n n 2 2 n n 2 n n2 n n n 3 1 3 3n+1 3 1 3 3 â&#x2C6;&#x2014; + 1 + 2 Â· n(n + 1) = 1+ f (xk )â&#x2C6;&#x2020;x = k = 2 n n2 2 n 2 2 2 n k=1 k=1 k=1 3 1 3 3 15 3 1+ 1+ = 1+ = A = lim nâ&#x2020;&#x2019;+â&#x2C6;&#x17E; 2 2 n 2 2 4

uh a

5 â&#x2C6;&#x2014; 5 25 25 5 5 â&#x2C6;&#x2014; â&#x2C6;&#x2014; 38. â&#x2C6;&#x2020;x = , xk = 0 + k ; f (xk )â&#x2C6;&#x2020;x = (5 â&#x2C6;&#x2019; xk )â&#x2C6;&#x2020;x = 5 â&#x2C6;&#x2019; k = â&#x2C6;&#x2019; 2k n n n n n n n n n 25 25 25 n + 1 25 1 â&#x2C6;&#x2019; 2 f (xâ&#x2C6;&#x2014;k )â&#x2C6;&#x2020;x = k = 25 â&#x2C6;&#x2019; 2 Â· n(n + 1) = 25 â&#x2C6;&#x2019; n n n 2 2 n k=1 k=1 k=1 1 25 25 25 A = lim 25 â&#x2C6;&#x2019; 1+ = 25 â&#x2C6;&#x2019; = nâ&#x2020;&#x2019;+â&#x2C6;&#x17E; 2 n 2 2

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Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Exercise Set 6.4

3 â&#x2C6;&#x2014; 3 k2 3 , xk = 0 + k ; f (xâ&#x2C6;&#x2014;k )â&#x2C6;&#x2020;x = 9 â&#x2C6;&#x2019; 9 2 n n n n n n n n 2 27 k 3 k2 27 2 â&#x2C6;&#x2014; = 9â&#x2C6;&#x2019;9 2 1 â&#x2C6;&#x2019; 2 = 27 â&#x2C6;&#x2019; 3 f (xk )â&#x2C6;&#x2020;x = k n n n n n k=1 k=1 k=1 k=1 n 27 2 1 = 18 A = lim 27 â&#x2C6;&#x2019; 3 k = 27 â&#x2C6;&#x2019; 27 nâ&#x2020;&#x2019;+â&#x2C6;&#x17E; n 3

237

us uf

39. â&#x2C6;&#x2020;x =

k=1

3 â&#x2C6;&#x2014; 3 ,x =k n k n 1 9k 2 3 1 â&#x2C6;&#x2014; 2 12 27k 2 â&#x2C6;&#x2014; f (xk )â&#x2C6;&#x2020;x = 4 â&#x2C6;&#x2019; (xk ) â&#x2C6;&#x2020;x = 4 â&#x2C6;&#x2019; = â&#x2C6;&#x2019; 4 4 n2 n n 4n3 n n n 27 2 12 â&#x2C6;&#x2019; 3 k f (xâ&#x2C6;&#x2014;k )â&#x2C6;&#x2020;x = 4n n k=1

k=1

40. â&#x2C6;&#x2020;x =

27 1 9 (n + 1)(2n + 1) = 12 â&#x2C6;&#x2019; 3 Â· n(n + 1)(2n + 1) = 12 â&#x2C6;&#x2019; 4n 6 8 n2 1 1 9 9 1+ 2+ = 12 â&#x2C6;&#x2019; (1)(2) = 39/4 A = lim 12 â&#x2C6;&#x2019; nâ&#x2020;&#x2019;+â&#x2C6;&#x17E; 8 n n 8

41. â&#x2C6;&#x2020;x =

4 â&#x2C6;&#x2014; 4 , xk = 2 + k n n

3 3 32 2 6 12 2 4 4 32 8 3 = 1+ k = 1 + k + 2k + 3k = = 2+ k n n n n n n n n n n n n n 32 6 12 2 8 3 f (xâ&#x2C6;&#x2014;k )â&#x2C6;&#x2020;x = 1+ k+ 2 k + 3 k n n n n k=1 k=1 k=1 k=1 k=1 32 6 1 12 1 8 1 2 2 = n + Â· n(n + 1) + 2 Â· n(n + 1)(2n + 1) + 3 Â· n (n + 1) n n 2 n 6 n 4 2 n+1 (n + 1)(2n + 1) (n + 1) = 32 1 + 3 +2 +2 2 n n n2 2 1 1 1 1 A = lim 32 1 + 3 1 + +2 1+ 2+ +2 1+ nâ&#x2020;&#x2019;+â&#x2C6;&#x17E; n n n n (xâ&#x2C6;&#x2014;k )3 â&#x2C6;&#x2020;x

f (xâ&#x2C6;&#x2014;k )â&#x2C6;&#x2020;x

= 32[1 + 3(1) + 2(1)(2) + 2(1)2 ] = 320

uh a

3 2 2 2 2 â&#x2C6;&#x2014; â&#x2C6;&#x2014; 3 â&#x2C6;&#x2014; 42. â&#x2C6;&#x2020;x = , xk = â&#x2C6;&#x2019;3 + k ; f (xk )â&#x2C6;&#x2020;x = [1 â&#x2C6;&#x2019; (xk ) ]â&#x2C6;&#x2020;x = 1 â&#x2C6;&#x2019; â&#x2C6;&#x2019;3 + k n n n n 54 8 3 2 36 2 28 â&#x2C6;&#x2019; k + 2 k â&#x2C6;&#x2019; 3 k = n n n n n (n + 1)(2n + 1) (n + 1)2 2 28n â&#x2C6;&#x2019; 27(n + 1) + 6 â&#x2C6;&#x2019;2 f (xâ&#x2C6;&#x2014;k )â&#x2C6;&#x2020;x = n n n k=1 2 1 1 1 1 A = lim 2 28 â&#x2C6;&#x2019; 27 1 + +6 1+ 2+ â&#x2C6;&#x2019;2 1+ nâ&#x2020;&#x2019;+â&#x2C6;&#x17E; n n n n = 2(28 â&#x2C6;&#x2019; 27 + 12 â&#x2C6;&#x2019; 2) = 22

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Chapter 6

3 â&#x2C6;&#x2014; 3 , x = 1 + (k â&#x2C6;&#x2019; 1) n k n 3 3 1 3 9 1 1 1 + (k â&#x2C6;&#x2019; 1) = + (k â&#x2C6;&#x2019; 1) 2 f (xâ&#x2C6;&#x2014;k )â&#x2C6;&#x2020;x = xâ&#x2C6;&#x2014;k â&#x2C6;&#x2020;x = 2 n n 2 n n 2 n n n 9 9 1 3 9nâ&#x2C6;&#x2019;1 1 1 3 + 2 3 + 2 Â· (n â&#x2C6;&#x2019; 1)n = + (k â&#x2C6;&#x2019; 1) = f (xâ&#x2C6;&#x2014;k )â&#x2C6;&#x2020;x = n n 2 n 2 2 4 n 2 k=1 k=1 k=1 3 9 1 15 3 9 = + = + 1â&#x2C6;&#x2019; A = lim nâ&#x2020;&#x2019;+â&#x2C6;&#x17E; 2 n 4 4 2 4

us uf

5 5 â&#x2C6;&#x2014; , x = (k â&#x2C6;&#x2019; 1) n k n

5 25 25 5 = â&#x2C6;&#x2019; 2 (k â&#x2C6;&#x2019; 1) f (xâ&#x2C6;&#x2014;k )â&#x2C6;&#x2020;x = (5 â&#x2C6;&#x2019; xâ&#x2C6;&#x2014;k )â&#x2C6;&#x2020;x = 5 â&#x2C6;&#x2019; (k â&#x2C6;&#x2019; 1) n n n n n n n 25 25 25 n â&#x2C6;&#x2019; 1 f (xâ&#x2C6;&#x2014;k )â&#x2C6;&#x2020;x = 1â&#x2C6;&#x2019; 2 (k â&#x2C6;&#x2019; 1) = 25 â&#x2C6;&#x2019; n n 2 n k=1 k=1 k=1 25 1 25 25 = 25 â&#x2C6;&#x2019; A = lim 25 â&#x2C6;&#x2019; 1â&#x2C6;&#x2019; = nâ&#x2020;&#x2019;+â&#x2C6;&#x17E; n 2 2 2

44. â&#x2C6;&#x2020;x =

43. â&#x2C6;&#x2020;x =

3 â&#x2C6;&#x2014; 3 (k â&#x2C6;&#x2019; 1)2 3 , xk = 0 + (k â&#x2C6;&#x2019; 1) ; f (xâ&#x2C6;&#x2014;k )â&#x2C6;&#x2020;x = (9 â&#x2C6;&#x2019; 9 ) n2 n n n n n n n n 2 (k â&#x2C6;&#x2019; 1) 3 (k â&#x2C6;&#x2019; 1)2 27 2 54 27 27 â&#x2C6;&#x2014; 9â&#x2C6;&#x2019;9 1 â&#x2C6;&#x2019; = 27 â&#x2C6;&#x2019; f (xk )â&#x2C6;&#x2020;x = k + kâ&#x2C6;&#x2019; 2 = 2 2 3 3 n n n n n n n k=1 k=1 k=1 k=1 k=1 1 + 0 + 0 = 18 A = lim = 27 â&#x2C6;&#x2019; 27 nâ&#x2020;&#x2019;+â&#x2C6;&#x17E; 3

45. â&#x2C6;&#x2020;x =

3 â&#x2C6;&#x2014; 3 , x = (k â&#x2C6;&#x2019; 1) n k n 1 9(k â&#x2C6;&#x2019; 1)2 3 1 â&#x2C6;&#x2014; 2 27k 27 12 27k 2 â&#x2C6;&#x2014; f (xk )â&#x2C6;&#x2020;x = 4 â&#x2C6;&#x2019; (xk ) â&#x2C6;&#x2020;x = 4 â&#x2C6;&#x2019; + 3â&#x2C6;&#x2019; 3 = â&#x2C6;&#x2019; 2 3 4 4 n n n 4n 2n 4n n n n n n 27 12 27 27 â&#x2C6;&#x2019; 3 f (xâ&#x2C6;&#x2014;k )â&#x2C6;&#x2020;x = k2 + 3 kâ&#x2C6;&#x2019; 3 1 n 4n 2n 4n

k=1

46. â&#x2C6;&#x2020;x =

k=1

27 27 n(n + 1) 27 1 â&#x2C6;&#x2019; 2 = 12 â&#x2C6;&#x2019; 3 Â· n(n + 1)(2n + 1) + 3 2n 2 4n 4n 6

1 â&#x2C6;&#x2014; 2k â&#x2C6;&#x2019; 1 , xk = n 2n

47. â&#x2C6;&#x2020;x =

27 9 (n + 1)(2n + 1) 27 27 + = 12 â&#x2C6;&#x2019; + â&#x2C6;&#x2019; 2 8 n2 4n 4n2 4n 1 1 9 9 1+ 2+ + 0 + 0 â&#x2C6;&#x2019; 0 = 12 â&#x2C6;&#x2019; (1)(2) = 39/4 A = lim 12 â&#x2C6;&#x2019; nâ&#x2020;&#x2019;+â&#x2C6;&#x17E; 8 n n 8

uh a

f (xâ&#x2C6;&#x2014;k )â&#x2C6;&#x2020;x = n

k2 (2k â&#x2C6;&#x2019; 1)2 1 k 1 = â&#x2C6;&#x2019; 3+ 3 2 3 (2n) n n n 4n

f (xâ&#x2C6;&#x2014;k )â&#x2C6;&#x2020;x =

k=1

n n n 1 2 1 1 k â&#x2C6;&#x2019; k + 1 n3 n3 4n3 k=1

k=1

Using Theorem 6.4.4, n 1 1 A = lim f (xâ&#x2C6;&#x2014;k )â&#x2C6;&#x2020;x = + 0 + 0 = nâ&#x2020;&#x2019;+â&#x2C6;&#x17E; 3 3 k=1

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Exercise Set 6.4

2k â&#x2C6;&#x2019; 1 2 â&#x2C6;&#x2014; , xk = â&#x2C6;&#x2019;1 + n n 2 2 2k â&#x2C6;&#x2019; 1 8k 2 2 8k 2 f (xâ&#x2C6;&#x2014;k )â&#x2C6;&#x2020;x = â&#x2C6;&#x2019;1 + = 3 â&#x2C6;&#x2019; 3+ 3â&#x2C6;&#x2019; n n n n n n

239

k=1

n n 8 2 8 2 k â&#x2C6;&#x2019; k+ 2 â&#x2C6;&#x2019;2 n3 n3 n k=1

A = lim

nâ&#x2020;&#x2019;+â&#x2C6;&#x17E;

us uf

f (xâ&#x2C6;&#x2014;k )â&#x2C6;&#x2020;x =

k=1

f (xâ&#x2C6;&#x2014;k )â&#x2C6;&#x2020;x =

k=1

8 2 +0+0â&#x2C6;&#x2019;2= 3 3

2k 2 â&#x2C6;&#x2014; , x = â&#x2C6;&#x2019;1 + n k n 2 2k 2 k =â&#x2C6;&#x2019; +4 2 f (xâ&#x2C6;&#x2014;k )â&#x2C6;&#x2020;x = â&#x2C6;&#x2019;1 + n n n n n n 4 4 n(n + 1) 2 f (xâ&#x2C6;&#x2014;k )â&#x2C6;&#x2020;x = â&#x2C6;&#x2019;2 + 2 k = â&#x2C6;&#x2019;2 + 2 = â&#x2C6;&#x2019;2 + 2 + n n 2 n

k=1

49. â&#x2C6;&#x2020;x =

k=1

nâ&#x2020;&#x2019;+â&#x2C6;&#x17E;

f (xâ&#x2C6;&#x2014;k )â&#x2C6;&#x2020;x = 0

A = lim

48. â&#x2C6;&#x2020;x =

k=1

3 â&#x2C6;&#x2014; 3k , x = â&#x2C6;&#x2019;1 + n k n 3 3k 3 9 â&#x2C6;&#x2014; f (xk )â&#x2C6;&#x2020;x = â&#x2C6;&#x2019;1 + = â&#x2C6;&#x2019; + 2k n n n n n 9 n(n + 1) f (xâ&#x2C6;&#x2014;k )â&#x2C6;&#x2020;x = â&#x2C6;&#x2019;3 + 2 n 2

50. â&#x2C6;&#x2020;x =

k=1

nâ&#x2020;&#x2019;+â&#x2C6;&#x17E;

9 3 +0= 2 2

A = lim

The area below the x-axis cancels the area above the x-axis.

f (xâ&#x2C6;&#x2014;k )â&#x2C6;&#x2020;x = â&#x2C6;&#x2019;3 +

k=1

The area below the x-axis cancels the area above the x-axis that lies to the right of the line x = 1; 1+2 3 the remaining area is a trapezoid of width 1 and heights 1, 2, hence its area is = 2 2 2 â&#x2C6;&#x2014; 2k ,x = n k n 2 8k 2 2k 2 2 â&#x2C6;&#x2014; = 3 â&#x2C6;&#x2019; f (xk ) = â&#x2C6;&#x2019;1 n n n n f (xâ&#x2C6;&#x2014;k )â&#x2C6;&#x2020;x =

51. â&#x2C6;&#x2020;x =

k=1

uh a

A = lim

nâ&#x2020;&#x2019;+â&#x2C6;&#x17E;

k=1

n n 8 2 2 8 n(n + 1)(2n + 1) k â&#x2C6;&#x2019; 1= 3 â&#x2C6;&#x2019;2 n3 n n 6 k=1

f (xâ&#x2C6;&#x2014;k )â&#x2C6;&#x2020;x =

k=1

2 16 â&#x2C6;&#x2019;2= 6 3

2 â&#x2C6;&#x2014; 2k , x = â&#x2C6;&#x2019;1 + n k n 3 2 2k k2 k3 2 k f (xâ&#x2C6;&#x2014;k )â&#x2C6;&#x2020;x = â&#x2C6;&#x2019;1 + = â&#x2C6;&#x2019; + 12 2 â&#x2C6;&#x2019; 24 3 + 16 4 n n n n n n

52. â&#x2C6;&#x2020;x =

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Chapter 6

f (xâ&#x2C6;&#x2014;k )â&#x2C6;&#x2020;x

k=1

12 n(n + 1) 24 n(n + 1)(2n + 1) 16 â&#x2C6;&#x2019; 3 + 4 = â&#x2C6;&#x2019;2 + 2 2 n 6 n n

A = lim

nâ&#x2020;&#x2019;+â&#x2C6;&#x17E;

f (xâ&#x2C6;&#x2014;k ) = â&#x2C6;&#x2019;2 +

k=1

n(n + 1) 2

12 48 16 â&#x2C6;&#x2019; + 2 =0 2 6 2

us uf

240

bâ&#x2C6;&#x2019;a bâ&#x2C6;&#x2019;a â&#x2C6;&#x2014; , xk = a + (k â&#x2C6;&#x2019; 1) n n bâ&#x2C6;&#x2019;a a bâ&#x2C6;&#x2019;a bâ&#x2C6;&#x2019;a (k â&#x2C6;&#x2019; 1) = m(b â&#x2C6;&#x2019; a) + (k â&#x2C6;&#x2019; 1) f (xâ&#x2C6;&#x2014;k )â&#x2C6;&#x2020;x = mxâ&#x2C6;&#x2014;k â&#x2C6;&#x2020;x = m a + n n n n2 n bâ&#x2C6;&#x2019;a nâ&#x2C6;&#x2019;1 Â· f (xâ&#x2C6;&#x2014;k )â&#x2C6;&#x2020;x = m(b â&#x2C6;&#x2019; a) a + n 2 k=1 1 b+a 1 bâ&#x2C6;&#x2019;a A = lim m(b â&#x2C6;&#x2019; a) a + 1â&#x2C6;&#x2019; = m(b â&#x2C6;&#x2019; a) = m(b2 â&#x2C6;&#x2019; a2 ) nâ&#x2020;&#x2019;+â&#x2C6;&#x17E; 2 n 2 2 bâ&#x2C6;&#x2019;a â&#x2C6;&#x2014; k , xk = a + (b â&#x2C6;&#x2019; a) n n mk ma â&#x2C6;&#x2014; f (xk )â&#x2C6;&#x2020;x = (b â&#x2C6;&#x2019; a) + 2 (b â&#x2C6;&#x2019; a)2 n n n n(n + 1) m f (xâ&#x2C6;&#x2014;k )â&#x2C6;&#x2020;x = ma(b â&#x2C6;&#x2019; a) + 2 (b â&#x2C6;&#x2019; a)2 n 2

53. â&#x2C6;&#x2020;x =

54. â&#x2C6;&#x2020;x =

A = lim

k=1

m a+b (b â&#x2C6;&#x2019; a)2 = m(b â&#x2C6;&#x2019; a) 2 2

nâ&#x2020;&#x2019;+â&#x2C6;&#x17E;

f (xâ&#x2C6;&#x2014;k )â&#x2C6;&#x2020;x = ma(b â&#x2C6;&#x2019; a) +

k=1

b â&#x2C6;&#x2014; b ,x = k n k n n n b4 b4 3 b4 (n + 1)2 f (xâ&#x2C6;&#x2014;k )â&#x2C6;&#x2020;x = (xâ&#x2C6;&#x2014;k )3 â&#x2C6;&#x2020;x = 4 k 3 , f (xâ&#x2C6;&#x2014;k )â&#x2C6;&#x2020;x = 4 k = n n 4 n2 k=1 k=1 2 1 b4 1+ = b4 /4 A = lim nâ&#x2020;&#x2019;+â&#x2C6;&#x17E; 4 n

55. (a) With xâ&#x2C6;&#x2014;k as the right endpoint, â&#x2C6;&#x2020;x =

uh a

bâ&#x2C6;&#x2019;a â&#x2C6;&#x2014; bâ&#x2C6;&#x2019;a , xk = a + k n n 3 bâ&#x2C6;&#x2019;a bâ&#x2C6;&#x2019;a k f (xâ&#x2C6;&#x2014;k )â&#x2C6;&#x2020;x = (xâ&#x2C6;&#x2014;k )3 â&#x2C6;&#x2020;x = a + n n 2 3a(b â&#x2C6;&#x2019; a)2 2 (b â&#x2C6;&#x2019; a)3 3 b â&#x2C6;&#x2019; a 3 3a (b â&#x2C6;&#x2019; a) a + k+ k + k = n n n2 n3 n (n + 1)(2n + 1) 3 n+1 1 â&#x2C6;&#x2014; f (xk )â&#x2C6;&#x2020;x = (b â&#x2C6;&#x2019; a) a3 + a2 (b â&#x2C6;&#x2019; a) + a(b â&#x2C6;&#x2019; a)2 2 n 2 n2 k=1 (n + 1)2 1 + (b â&#x2C6;&#x2019; a)3 4 n2 n A = lim f (xâ&#x2C6;&#x2014;k )â&#x2C6;&#x2020;x

(b) â&#x2C6;&#x2020;x =

nâ&#x2020;&#x2019;+â&#x2C6;&#x17E;

k=1

1 3 1 = (b â&#x2C6;&#x2019; a) a3 + a2 (b â&#x2C6;&#x2019; a) + a(b â&#x2C6;&#x2019; a)2 + (b â&#x2C6;&#x2019; a)3 = (b4 â&#x2C6;&#x2019; a4 ). 2 4 4

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Exercise Set 6.4

241

k =2Âˇ

k=1

m+1

if n = 2m + 1 then (2m + 1) + (2m â&#x2C6;&#x2019; 1) + Âˇ Âˇ Âˇ + 5 + 3 + 1 = kâ&#x2C6;&#x2019;

k=1

1=2Âˇ

k=1

(m + 1)(m + 2) n2 + 2n + 1 â&#x2C6;&#x2019; (m + 1) = (m + 1)2 = 2 4

58. 50Âˇ30+49Âˇ29+Âˇ Âˇ Âˇ+22Âˇ2+21Âˇ1 =

k(k+20) =

k=1

k 2 +20

k=1

59. both are valid

61.

k=1

60. none is valid

(ak â&#x2C6;&#x2019; bk ) = (a1 â&#x2C6;&#x2019; b1 ) + (a2 â&#x2C6;&#x2019; b2 ) + Âˇ Âˇ Âˇ + (an â&#x2C6;&#x2019; bn )

k=1

= (a1 + a2 + Âˇ Âˇ Âˇ + an ) â&#x2C6;&#x2019; (b1 + b2 + Âˇ Âˇ Âˇ + bn ) =

ak â&#x2C6;&#x2019;

k=1

quantity in brackets to get

n (k + 1)4 â&#x2C6;&#x2019; k 4 = (n + 1)4 â&#x2C6;&#x2019; 1 (telescoping sum), expand the

(4k 3 + 6k 2 + 4k + 1) = (n + 1)4 â&#x2C6;&#x2019; 1,

62.

30 Âˇ 31 Âˇ 61 30 Âˇ 31 +20 = 18,755 6 2

m+1

(2k â&#x2C6;&#x2019; 1)

k=1

m+1

n2 + 2n m(m + 1) = m(m + 1) = ; 2 4

57. If n = 2m then 2m + 2(m â&#x2C6;&#x2019; 1) + Âˇ Âˇ Âˇ + 2 Âˇ 2 + 2 = 2

us uf

56. Let A be the area of the region under the curve and above the interval 0 â&#x2030;¤ x â&#x2030;¤ 1 on the x-axis, and let B be the area of the region between the curve and the interval 0 â&#x2030;¤ y â&#x2030;¤ 1 on the y-axis. Together A and B form the square of side 1, so A + B = 1. But B can also be considered as the area between the curve x = y 2 and the interval 0 â&#x2030;¤ y â&#x2030;¤ 1 on 1 1 2 the y-axis. By Exercise 47 above, B = ,so A = 1 â&#x2C6;&#x2019; = . 3 3 3

k=1

k3 + 6

k=1

1 = (n + 1)4 â&#x2C6;&#x2019; 1

k=1 n

1 (n + 1)4 â&#x2C6;&#x2019; 1 â&#x2C6;&#x2019; 6 k = 4

k=1

k â&#x2C6;&#x2019;4

n k=1

kâ&#x2C6;&#x2019;

k=1

1 [(n + 1)4 â&#x2C6;&#x2019; 1 â&#x2C6;&#x2019; n(n + 1)(2n + 1) â&#x2C6;&#x2019; 2n(n + 1) â&#x2C6;&#x2019; n] 4

1 (n + 1)[(n + 1)3 â&#x2C6;&#x2019; n(2n + 1) â&#x2C6;&#x2019; 2n â&#x2C6;&#x2019; 1] 4

k=1

k2 + 4

1 1 (n + 1)(n3 + n2 ) = n2 (n + 1)2 4 4

uh a

63. (a)

1 means add 1 to itself n times, which gives the result.

k=1

(b)

n n 1 1 n(n + 1) 1 1 1 1 k = k= = + , so lim nâ&#x2020;&#x2019;+â&#x2C6;&#x17E; n2 n2 n2 2 2 2n 2 k=1

k=1

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Chapter 6

k=1

2 n n 1 3 1 3 1 n(n + 1) 1 1 1 1 (d) k = = , so lim k = + + nâ&#x2020;&#x2019;+â&#x2C6;&#x17E; n4 n4 n4 2 4 2n 4n2 4 k=1

k=1

EXERCISE SET 6.5 (b) 2

1. (a) (4/3)(1) + (5/2)(1) + (4)(2) = 71/6

(c)

n n 2 3 1 1 2 1 2 1 n(n + 1)(2n + 1) 1 = + + k = 3 , so lim k = nâ&#x2020;&#x2019;+â&#x2C6;&#x17E; n3 n 6 6 6n 6n2 3 n3

us uf

242

â&#x2C6;&#x161; â&#x2C6;&#x161; â&#x2C6;&#x161; 2. (a) ( 2/2)(Ï&#x20AC;/2) + (â&#x2C6;&#x2019;1)(3Ï&#x20AC;/4) + (0)(Ï&#x20AC;/2) + ( 2/2)(Ï&#x20AC;/4) = 3( 2 â&#x2C6;&#x2019; 2)Ï&#x20AC;/8 (b) 3Ï&#x20AC;/4

(b) 2

2 2

x dx

10. (a)

(b)

sin2 x dx

lim

max â&#x2C6;&#x2020;xk â&#x2020;&#x2019;0

lim

max â&#x2C6;&#x2020;xk â&#x2020;&#x2019;0

(b)

xâ&#x2C6;&#x2014;k â&#x2C6;&#x2020;xk ; a = 0, b = 1 +1

max â&#x2C6;&#x2020;xk â&#x2020;&#x2019;0 xâ&#x2C6;&#x2014; k=1 k

(1 + cos xâ&#x2C6;&#x2014;k ) â&#x2C6;&#x2020;xk , a = â&#x2C6;&#x2019;Ï&#x20AC;/2, b = Ï&#x20AC;/2

k=1

1 (3)(3) = 9/2 2

lim

xâ&#x2C6;&#x2014;k â&#x2C6;&#x2020;xk , a = 1, b = 2

k=1 n

lim

11. (a) A =

k=1 n

2xâ&#x2C6;&#x2014;k â&#x2C6;&#x2020;xk ; a = 1, b = 2

9. (a)

Ï&#x20AC;/2

â&#x2C6;&#x2019;3

4x(1 â&#x2C6;&#x2019; 3x)dx

1 (b) â&#x2C6;&#x2019;A = â&#x2C6;&#x2019; (1)(1 + 2) = â&#x2C6;&#x2019;3/2 2

x3 dx

â&#x2C6;&#x2019;1

4. (a) (â&#x2C6;&#x2019;8)(2) + (0)(1) + (0)(1) + (8)(2) = 0

3. (a) (â&#x2C6;&#x2019;9/4)(1) + (3)(2) + (63/16)(1) + (â&#x2C6;&#x2019;5)(3) = â&#x2C6;&#x2019;117/16 (b) 3

y x

-2 -1 A

uh a

1 (c) â&#x2C6;&#x2019;A1 + A2 = â&#x2C6;&#x2019; + 8 = 15/2 2

-1

(d) â&#x2C6;&#x2019;A1 + A2 = 0 y

-5 A2

x 5

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Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Exercise Set 6.5

1 (1)(2) = 1 2

12. (a) A =

1 (2)(3/2 + 1/2) = 2 2

(b) A =

us uf

-1

1 (c) −A = − (1/2)(1) = −1/4 2

(d) A1 − A2 = 1 − 1/4 = 3/4 y

y 1

x 2

(b) 0; A1 = A2 by symmetry

13. (a) A = 2(5) = 10 y

1 1 (5)(5/2) + (1)(1/2) 2 2 = 13/2

3 2

-1

A1 2

14. (a) A = (6)(5) = 30 y

1 [π(1)2 ] = π/2 2

(d)

1 y A x

-1

(b) −A1 + A2 = 0 because A1 = A2 by symmetry

-5

-10

1 1 (2)(2) + (1)(1) = 5/2 2 2

(d)

1 π(2)2 = π 4

y 2

uh a

15. (a) 0.8

243

x 2

(b) −2.6

(d) −0.3

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Chapter 6

2xdx = x

(b)

2xdx = x2

â&#x2C6;&#x2019;1

(c)

(d)

1/2

â&#x2C6;&#x2019;1

g(x)dx = 3(2) â&#x2C6;&#x2019; 10 = â&#x2C6;&#x2019;4 1

0 â&#x2C6;&#x2019;2

f (x)dx = â&#x2C6;&#x2019;

xdx + 2

â&#x2C6;&#x2019;1

(b) 4

dx â&#x2C6;&#x2019; 5

2dx + â&#x2C6;&#x2019;3

â&#x2C6;&#x2019;2

â&#x2C6;&#x161;

dx â&#x2C6;&#x2019; 3

â&#x2C6;&#x2019;2

f (x)dx = â&#x2C6;&#x2019;(2 â&#x2C6;&#x2019; 6) = 4 1

1 â&#x2C6;&#x2019; x2 dx = 1/2 + 2(Ï&#x20AC;/4) = (1 + Ï&#x20AC;)/2

xdx = 4 Â· 4 â&#x2C6;&#x2019; 5(â&#x2C6;&#x2019;1/2 + (3 Â· 3)/2) = â&#x2C6;&#x2019;4

f (x)dx + â&#x2C6;&#x2019;2

â&#x2C6;&#x2019;3

(b)

â&#x2C6;&#x2019;1

22. (a)

f (x)dx = â&#x2C6;&#x2019;

â&#x2C6;&#x2019;2

21. (a)

f (x)dx = 1 â&#x2C6;&#x2019; (â&#x2C6;&#x2019;2) = 3 0

f (x)dx â&#x2C6;&#x2019;

f (x)dx =

23. (a)

1/2

f (x)dx â&#x2C6;&#x2019;

20.

+ 2xï£» = 12 â&#x2C6;&#x2019;(1/2)2 +2Â·5â&#x2C6;&#x2019;2Â·1 = 3/4+8 = 35/4

2dx = x2

g(x)dx = 5 + 2(â&#x2C6;&#x2019;3) = â&#x2C6;&#x2019;1

â&#x2C6;&#x2019;1

18. 3

ï£¹5

f (x)dx + 2

19.

2xdx +

1/2

f (x)dx =

17.

â&#x2C6;&#x2019;1

10 2dx = 2x = 18

= 12 â&#x2C6;&#x2019; (â&#x2C6;&#x2019;1)2 = 0

f (x)dx = 1

0 1

f (x)dx =

us uf

f (x)dx =

16. (a)

9 â&#x2C6;&#x2019; x2 dx = 2 Â· 3 + (Ï&#x20AC;(3)2 )/4 = 6 + 9Ï&#x20AC;/4

244

|x|dx = 4 Â· 1 â&#x2C6;&#x2019; 3(2)(2 Â· 2)/2 = â&#x2C6;&#x2019;8

x > 0, 1 â&#x2C6;&#x2019; x < 0 on [2, 3] so the integral is negative

24. (a) x4 > 0,

â&#x2C6;&#x161;

(b) x > 0, 3 â&#x2C6;&#x2019; cos x > 0 for all x so the integral is positive 2

3 â&#x2C6;&#x2019; x > 0 on [â&#x2C6;&#x2019;3, â&#x2C6;&#x2019;1] so the integral is positive

27.

25 â&#x2C6;&#x2019; (x â&#x2C6;&#x2019;

25.

(b) x â&#x2C6;&#x2019; 9 < 0, |x| + 1 > 0 on [â&#x2C6;&#x2019;2, 2] so the integral is negative 3

5)2 dx

uh a

(3x + 1)dx = 5/2

= Ï&#x20AC;(5) /2 = 25Ï&#x20AC;/2

26.

9 â&#x2C6;&#x2019; (x â&#x2C6;&#x2019; 3)2 dx = Ï&#x20AC;(3)2 /4 = 9Ï&#x20AC;/4

28. â&#x2C6;&#x2019;2

4 â&#x2C6;&#x2019; x2 dx = Ï&#x20AC;(2)2 /2 = 2Ï&#x20AC;

29. (a) f is continuous on [â&#x2C6;&#x2019;1, 1] so f is integrable there by Part (a) of Theorem 6.5.8 (b) |f (x)| â&#x2030;¤ 1 so f is bounded on [â&#x2C6;&#x2019;1, 1], and f has one point of discontinuity, so by Part (b) of Theorem 6.5.8 f is integrable on [â&#x2C6;&#x2019;1, 1]

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Exercise Set 6.5

245

1 does not exist. f is continuous x elsewhere. â&#x2C6;&#x2019;1 â&#x2030;¤ f (x) â&#x2030;¤ 1 for x in [â&#x2C6;&#x2019;1, 1] so f is bounded there. By Part (b), Theorem 6.5.8, f is integrable on [â&#x2C6;&#x2019;1, 1].

(d) f (x) is discontinuous at the point x = 0 because lim sin

us uf

xâ&#x2020;&#x2019;0

30. Each subinterval of a partition of [a, b] contains both rational and irrational numbers. If all xâ&#x2C6;&#x2014;k are chosen to be rational then n n n n f (xâ&#x2C6;&#x2014;k )â&#x2C6;&#x2020;xk = (1)â&#x2C6;&#x2020;xk = â&#x2C6;&#x2020;xk = b â&#x2C6;&#x2019; a so lim f (xâ&#x2C6;&#x2014;k )â&#x2C6;&#x2020;xk = b â&#x2C6;&#x2019; a. xâ&#x2C6;&#x2014;k

max â&#x2C6;&#x2020;xk â&#x2020;&#x2019;0

k=1

are irrational then

lim

max â&#x2C6;&#x2020;xk â&#x2020;&#x2019;0

the preceding limits are not equal. 31. (a) Let Sn = that

n k=1

max â&#x2C6;&#x2020;xk â&#x2020;&#x2019;0

k=1

= 0. Thus f is not integrable on [a, b] because

k=1

f (xâ&#x2C6;&#x2014;k )â&#x2C6;&#x2020;xk and S =

lim

f (xâ&#x2C6;&#x2014;k )â&#x2C6;&#x2020;xk

If all

k=1

f (x)dx then a

n k=1

cf (xâ&#x2C6;&#x2014;k )â&#x2C6;&#x2020;xk = cSn and we want to prove

k=1

cSn = cS. If c = 0 the result follows immediately, so suppose that c = 0 then

for any " > 0, |cSn â&#x2C6;&#x2019; cS| = |c||Sn â&#x2C6;&#x2019; S| < " if |Sn â&#x2C6;&#x2019; S| < "/|c|. But because f is integrable on [a, b], there is a number Î´ > 0 such that |Sn â&#x2C6;&#x2019; S| < "/|c| whenever max â&#x2C6;&#x2020;xk < Î´ so |cSn â&#x2C6;&#x2019; cS| < " and hence lim cSn = cS. max â&#x2C6;&#x2020;xk â&#x2020;&#x2019;0 b n n n â&#x2C6;&#x2014; â&#x2C6;&#x2014; â&#x2C6;&#x2014; â&#x2C6;&#x2014; (b) Let Rn = f (xk )â&#x2C6;&#x2020;xk , Sn = g(xk )â&#x2C6;&#x2020;xk , Tn = [f (xk ) + g(xk )]â&#x2C6;&#x2020;xk , R = f (x)dx,

k=1 k=1 k=1 b and S = g(x)dx then Tn = Rn + Sn and we want to prove that a

lim

max â&#x2C6;&#x2020;xk â&#x2020;&#x2019;0

Tn = R + S.

max â&#x2C6;&#x2020;xk â&#x2020;&#x2019;0

32. For the smallest, ďŹ nd xâ&#x2C6;&#x2014;k so that f (xâ&#x2C6;&#x2014;k ) is minimum on each subinterval: xâ&#x2C6;&#x2014;1 = 1, xâ&#x2C6;&#x2014;2 = 3/2, xâ&#x2C6;&#x2014;3 = 3 so (2)(1) + (7/4)(2) + (4)(1) = 9.5. For the largest, ďŹ nd xâ&#x2C6;&#x2014;k so that f (xâ&#x2C6;&#x2014;k ) is maximum on each subinterval: xâ&#x2C6;&#x2014;1 = 0, xâ&#x2C6;&#x2014;2 = 3, xâ&#x2C6;&#x2014;3 = 4 so (4)(1) + (4)(2) + (8)(1) = 20. 4k 2 4(k â&#x2C6;&#x2019; 1)2 4 4k 2 â&#x2C6;&#x2014; â&#x2C6;&#x2019; = (2k â&#x2C6;&#x2019; 1), x = , k n2 n2 n2 n2 2k 8k 8 , f (xâ&#x2C6;&#x2014;k )â&#x2C6;&#x2020;xk = 3 (2k â&#x2C6;&#x2019; 1) = 3 (2k 2 â&#x2C6;&#x2019; k), f (xâ&#x2C6;&#x2014;k ) = n n n n n 1 4 (n + 1)(4n â&#x2C6;&#x2019; 1) 1 8 8 n(n + 1)(2n + 1) â&#x2C6;&#x2019; n(n + 1) = f (xâ&#x2C6;&#x2014;k )â&#x2C6;&#x2020;xk = 3 (2k 2 â&#x2C6;&#x2019; k) = 3 , n n 3 2 3 n2 k=1 k=1 n 1 1 16 4 â&#x2C6;&#x2014; 1+ 4â&#x2C6;&#x2019; = . f (xk )â&#x2C6;&#x2020;xk = lim lim nâ&#x2020;&#x2019;+â&#x2C6;&#x17E; nâ&#x2020;&#x2019;+â&#x2C6;&#x17E; 3 n n 3

uh a

33. â&#x2C6;&#x2020;xk =

k=1

34. For any partition of [a, b] use the right endpoints to form the sum

f (x) dx = lim

for each k, the sum is zero and so is a

nâ&#x2020;&#x2019;+â&#x2C6;&#x17E;

f (xâ&#x2C6;&#x2014;k )â&#x2C6;&#x2020;xk . Since f (xâ&#x2C6;&#x2014;k ) = 0

k=1

f (xâ&#x2C6;&#x2014;k )â&#x2C6;&#x2020;xk .

k=1

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246

Chapter 6

35. With f (x) = g(x) then f (x) â&#x2C6;&#x2019; g(x) = 0 for a < x â&#x2030;¤ b. By Theorem 6.5.4(b) b b b b f (x) dx = [(f (x) â&#x2C6;&#x2019; g(x) + g(x)]dx = [f (x) â&#x2C6;&#x2019; g(x)]dx + g(x)dx. a

us uf

But the ďŹ rst term on the right hand side is zero (from Exercise 34), so b b f (x) dx = g(x) dx a

36. Choose any large positive integer N and any partition of [0, a]. Then choose xâ&#x2C6;&#x2014;1 in the ďŹ rst interval so small that f (xâ&#x2C6;&#x2014;1 )â&#x2C6;&#x2020;x1 > N . For example choose xâ&#x2C6;&#x2014;1 < â&#x2C6;&#x2020;x1 /N . Then with this partition and n f (xâ&#x2C6;&#x2014;k )â&#x2C6;&#x2020;xk > f (xâ&#x2C6;&#x2014;1 )â&#x2C6;&#x2020;x1 > N . This shows that the sum is dependent on partition choice of xâ&#x2C6;&#x2014;1 , k=1

and/or points, so DeďŹ nition 6.5.1 is not satisďŹ ed.

2 (2 â&#x2C6;&#x2019; x)dx = (2x â&#x2C6;&#x2019; x2 /2) = 4 â&#x2C6;&#x2019; 4/2 = 2 0

(b) â&#x2C6;&#x2019;1 3

1 2dx = 2x = 2(1) â&#x2C6;&#x2019; 2(â&#x2C6;&#x2019;1) = 4 â&#x2C6;&#x2019;1

3 (x + 1)dx = (x2 /2 + x) = 9/2 + 3 â&#x2C6;&#x2019; (1/2 + 1) = 6

(c)

xdx = x2 /2

2. (a) 0

â&#x2C6;&#x2019;1

= 81/4 â&#x2C6;&#x2019; 16/4 = 65/4

2 xdx = x3/2 3 3

ex dx = ex 1

uh a

xâ&#x2C6;&#x2019;2 dx = â&#x2C6;&#x2019;

4 5/2 x 5

1 x

x4 dx = x5 /5

â&#x2C6;&#x2019;3/5

8. 1

10.

5 dx = x2/5 2

14.

= 1

5 2/5 (4 â&#x2C6;&#x2019; 1) 2

2 = 81/10 â&#x2C6;&#x2019;1

xâ&#x2C6;&#x2019;6 dx = â&#x2C6;&#x2019;

5 1 dx = ln x = ln 5 â&#x2C6;&#x2019; ln 1 = ln 5 x 1

1 2 1 5 x + x 2 5

12.

= 1/5 â&#x2C6;&#x2019; (â&#x2C6;&#x2019;1)/5 = 2/5 â&#x2C6;&#x2019;1

= 2/3

9 4

â&#x2C6;&#x2019;1

= 844/5

0 1 3 2 x â&#x2C6;&#x2019; 2x + 7x = 48 3 â&#x2C6;&#x2019;3

11.

13.

= e3 â&#x2C6;&#x2019; e

2 = (27 â&#x2C6;&#x2019; 1) = 52/3 3

â&#x2C6;&#x161;

9 5dx = 5x = 5(9) â&#x2C6;&#x2019; 5(3) = 30

â&#x2C6;&#x2019;1

x3 dx = x4 /4

2 2 (x + 3)dx = (x /2 + 3x) = 4/2 + 6 â&#x2C6;&#x2019; (1/2 â&#x2C6;&#x2019; 3) = 21/2

(b)

(c)

= 25/2

1. (a)

EXERCISE SET 6.6

3x5/3 +

4 x

1 5x5

2 = 31/160 1

8 = 179/2 1

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Exercise Set 6.6

16.

17.

Ï&#x20AC;/4 â&#x2C6;&#x161; sin x = 2

18.

19.

5ex

â&#x2C6;&#x2019;Ï&#x20AC;/2

â&#x2C6;&#x2019;Ï&#x20AC;/4

3 ln 2

tan Î¸

= 5e3 â&#x2C6;&#x2019; 5(2) = 5e3 â&#x2C6;&#x2019; 10

20.

Ï&#x20AC;/4 0

1 1 2 x â&#x2C6;&#x2019; sec x = 3/2 â&#x2C6;&#x2019; sec(1) 2 0 1

(ln x)/2

1/2

= (ln 2)/2

1/â&#x2C6;&#x161;2 â&#x2C6;&#x161; x = sinâ&#x2C6;&#x2019;1 (1/ 2) â&#x2C6;&#x2019; sinâ&#x2C6;&#x2019;1 0 = Ï&#x20AC;/4

â&#x2C6;&#x2019;1

21. sin

â&#x2C6;&#x2019;1

22. tan

1 x = tanâ&#x2C6;&#x2019;1 1 â&#x2C6;&#x2019; tanâ&#x2C6;&#x2019;1 (â&#x2C6;&#x2019;1) = Ï&#x20AC;/4 â&#x2C6;&#x2019; (â&#x2C6;&#x2019;Ï&#x20AC;/4) = Ï&#x20AC;/2 â&#x2C6;&#x2019;1

2 â&#x2C6;&#x161; x â&#x2C6;&#x161; = secâ&#x2C6;&#x2019;1 2 â&#x2C6;&#x2019; secâ&#x2C6;&#x2019;1 2 = Ï&#x20AC;/3 â&#x2C6;&#x2019; Ï&#x20AC;/4 = Ï&#x20AC;/12

23. sec

â&#x2C6;&#x2019;1

4 = â&#x2C6;&#x2019;55/3 1

Ï&#x20AC;/2 â&#x2C6;&#x161; 1 2 x â&#x2C6;&#x2019; 2 cot x = Ï&#x20AC; 2 /9 + 2 3 2 Ï&#x20AC;/6

3/2

(3 â&#x2C6;&#x2019; 2x)dx +

29. (a) 0

26.

28.

3/2

Ï&#x20AC;/2

(b)

3Ï&#x20AC;/4

cos x dx + 0

4 2 â&#x2C6;&#x161; 8 y + y 3/2 â&#x2C6;&#x2019; 3/2 3 3y

1/2

2 x â&#x2C6;&#x2019; x3/2 3

4a a

3/2 2 (2x â&#x2C6;&#x2019; 3)dx = (3x â&#x2C6;&#x2019; x2 ) + (x2 â&#x2C6;&#x2019; 3x)

9 = 10819/324 4

5 = â&#x2C6;&#x2019; a3/2 3

= 9/4 + 1/4 = 5/2

3/2

Ï&#x20AC;/2 3Ï&#x20AC;/4 â&#x2C6;&#x161; (â&#x2C6;&#x2019; cos x)dx = sin x â&#x2C6;&#x2019; sin x = 2 â&#x2C6;&#x2019; 2/2

â&#x2C6;&#x161; 10 2 6 t â&#x2C6;&#x2019; t3/2 + â&#x2C6;&#x161; 3 t

27.

â&#x2C6;&#x2019; 2

25.

â&#x2C6;&#x2019;2/â&#x2C6;&#x161;3 â&#x2C6;&#x161; â&#x2C6;&#x161; = â&#x2C6;&#x2019; secâ&#x2C6;&#x2019;1 (â&#x2C6;&#x2019;2/ 3) + secâ&#x2C6;&#x2019;1 (â&#x2C6;&#x2019; 2) = â&#x2C6;&#x2019;5Ï&#x20AC;/6 + 3Ï&#x20AC;/4 = â&#x2C6;&#x2019;Ï&#x20AC;/12 x â&#x2C6;&#x161;

24. â&#x2C6;&#x2019; sec

â&#x2C6;&#x2019;1

Ï&#x20AC;/2

0 2 2 2 3/2 3/2 30. (a) 2 â&#x2C6;&#x2019; x dx + 2 + x dx = â&#x2C6;&#x2019; (2 â&#x2C6;&#x2019; x) + (2 + x) 3 3 â&#x2C6;&#x2019;1 0 â&#x2C6;&#x2019;1 0 â&#x2C6;&#x161; â&#x2C6;&#x161; â&#x2C6;&#x161; â&#x2C6;&#x161; 2 â&#x2C6;&#x161; 2 2 = â&#x2C6;&#x2019; (2 2 â&#x2C6;&#x2019; 3 3) + (8 â&#x2C6;&#x2019; 2 2) = (8 â&#x2C6;&#x2019; 4 2 + 3 3) 3 3 3 Ï&#x20AC;/2 Ï&#x20AC;/6 (1/2 â&#x2C6;&#x2019; sin x) dx + (sin x â&#x2C6;&#x2019; 1/2) dx (b) â&#x2C6;&#x161;

â&#x2C6;&#x161;

uh a

Ï&#x20AC;/6

Ï&#x20AC;/6 Ï&#x20AC;/2 = (x/2 + cos x) â&#x2C6;&#x2019; (cos x + x/2) = (Ï&#x20AC;/12 +

â&#x2C6;&#x161;

Ï&#x20AC;/6

â&#x2C6;&#x161; â&#x2C6;&#x161; 3/2) â&#x2C6;&#x2019; 1 â&#x2C6;&#x2019; Ï&#x20AC;/4 + ( 3/2 + Ï&#x20AC;/12) = 3 â&#x2C6;&#x2019; Ï&#x20AC;/12 â&#x2C6;&#x2019; 1

us uf

â&#x2C6;&#x2019; cos Î¸

Ï&#x20AC;/2

15.

247

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Chapter 6

(1â&#x2C6;&#x2019;e )dx+

31. (a) â&#x2C6;&#x2019;1

(b) 1

â&#x2C6;&#x2019;1

2â&#x2C6;&#x2019;x dx + x

0 1 x (e â&#x2C6;&#x2019;1)dx = (xâ&#x2C6;&#x2019;e ) +(e â&#x2C6;&#x2019;x) = â&#x2C6;&#x2019;1â&#x2C6;&#x2019;(â&#x2C6;&#x2019;1â&#x2C6;&#x2019;eâ&#x2C6;&#x2019;1 )+eâ&#x2C6;&#x2019;1â&#x2C6;&#x2019;1 = e+1/eâ&#x2C6;&#x2019;2 x

15 is an even function and changes sign at x = 2, thus +1 3 2 3 |f (x)| dx = 2 |f (x)| dx = â&#x2C6;&#x2019;2 f (x) dx + 2 f (x) dx

32. (a) The function f (x) = x2 â&#x2C6;&#x2019; 1 â&#x2C6;&#x2019; 3

â&#x2C6;&#x2019;3

2 4 xâ&#x2C6;&#x2019;2 dx = 2 ln x â&#x2C6;&#x2019; 1 + 2 â&#x2C6;&#x2019; 2 ln x = 2 ln 2 + 1 â&#x2C6;&#x2019; 2 ln 4 + 2 ln 2 = 1 x 1 2

us uf

248

28 â&#x2C6;&#x2019; 30 tanâ&#x2C6;&#x2019;1 (3) + 60 tanâ&#x2C6;&#x2019;1 (2) 3 â&#x2C6;&#x161;3/2 â&#x2C6;&#x161;2/2 â&#x2C6;&#x161;3/2 â&#x2C6;&#x161; â&#x2C6;&#x161; â&#x2C6;&#x161; 1 1 1 â&#x2C6;&#x161; â&#x2C6;&#x161; â&#x2C6;&#x161; (b) â&#x2C6;&#x2019; 2 dx â&#x2C6;&#x2019; 2 dx = â&#x2C6;&#x2019; â&#x2C6;&#x2019; 2 dx + â&#x2C6;&#x161; 1 â&#x2C6;&#x2019; x2 1 â&#x2C6;&#x2019; x2 1 â&#x2C6;&#x2019; x2 0 2/2 0 â&#x2C6;&#x161; â&#x2C6;&#x161; â&#x2C6;&#x161; â&#x2C6;&#x161; â&#x2C6;&#x161; â&#x2C6;&#x161; Ï&#x20AC; Ï&#x20AC; 2 3 3 2 3 â&#x2C6;&#x2019;1 â&#x2C6;&#x2019;1 + sin â&#x2C6;&#x2019; 2 + 1 = â&#x2C6;&#x2019;2 + â&#x2C6;&#x2019; â&#x2C6;&#x161; + 2 = â&#x2C6;&#x2019;2 sin â&#x2C6;&#x2019; 2 2 2 2 4 3 2 â&#x2C6;&#x161; 3 Ï&#x20AC; =2â&#x2C6;&#x2019; â&#x2C6;&#x161; â&#x2C6;&#x2019; 6 2

ï£± 1 2 ï£´ ï£´ xâ&#x2030;¤1 ï£² x , 2 (b) F (x) = ï£´ ï£´ 1 x3 + 1 , x > 1 ï£³ 6 3

â&#x2C6;&#x161;

x dx +

1 2 dx = x3/2 2 x 3

â&#x2C6;&#x2019;

1 x

= 17/12

34. (a)

33. (a) 17/6

ï£± 2 ï£´ ï£² x3/2 , x<1 3 (b) F (x) = ï£´ ï£³ â&#x2C6;&#x2019;1 + 5, x â&#x2030;¥ 1 x 3 3

35. 0.665867079; 1

1 1 dx = â&#x2C6;&#x2019; x2 x

= 2/3

36. 1.000257067;

Ï&#x20AC;/2

Ï&#x20AC;/2 sin x dx = â&#x2C6;&#x2019; cos x =1

1 37. 3.106017890; sec2 x dx = tan x = 2 tan 1 â&#x2030;&#x2C6; 3.114815450 â&#x2C6;&#x2019;1

uh a

38. 1.098242635;

39.

3 1 dx = ln x = ln 3 â&#x2030;&#x2C6; 1.098612289 x 1

(x + 1)dx =

40.

â&#x2C6;&#x2019;1

3 1 3 = 12 x +x 3 0

(â&#x2C6;&#x2019;x + 3x â&#x2C6;&#x2019; 2)dx =

A= 1

2 1 3 3 2 â&#x2C6;&#x2019; x + x â&#x2C6;&#x2019; 2x = 1/6 3 2 1

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Exercise Set 6.6

41.

2Ď&#x20AC;/3

3 sin x dx = â&#x2C6;&#x2019;3 cos x

A= 0

= 9/2

A=â&#x2C6;&#x2019;

42.

â&#x2C6;&#x2019;1

â&#x2C6;&#x2019;2

1 x dx = â&#x2C6;&#x2019; x4 4

â&#x2C6;&#x2019;1

= 15/4 â&#x2C6;&#x2019;2

0.8 1 â&#x2C6;&#x2019;1 â&#x2C6;&#x161; 43. (a) A = dx = sin x = sinâ&#x2C6;&#x2019;1 (0.8) 1 â&#x2C6;&#x2019; x2 0 0 (b) The calculator was in degree mode instead of radian mode; the correct answer is 0.93. 0.8

us uf

249

44. (a) the area is positive 5 5 1 1 4 1 3 1 1 1 1 1 343 (b) x â&#x2C6;&#x2019; x2 â&#x2C6;&#x2019; x + dx = x â&#x2C6;&#x2019; x3 â&#x2C6;&#x2019; x2 + x = 100 20 25 5 400 60 50 5 1200 â&#x2C6;&#x2019;2 â&#x2C6;&#x2019;2

â&#x2C6;&#x2019;Ď&#x20AC;/2

45. (a) the area between the curve and the x-axis breaks into equal parts, one above and one below the x-axis, so the integral is zero 1 1 1 1 (b) x3 dx = x4 = (14 â&#x2C6;&#x2019; (â&#x2C6;&#x2019;1)4 ) = 0; 4 4 â&#x2C6;&#x2019;1 â&#x2C6;&#x2019;1 Ď&#x20AC;/2 Ď&#x20AC;/2 sin xdx = â&#x2C6;&#x2019; cos x = â&#x2C6;&#x2019; cos(Ď&#x20AC;/2) + cos(â&#x2C6;&#x2019;Ď&#x20AC;/2) = 0 + 0 = 0

â&#x2C6;&#x2019;Ď&#x20AC;/2

(c) The a area on the a left side of the y-axis is equal to the area on the right side, so f (x)dx = 2 f (x)dx â&#x2C6;&#x2019;a

x2 dx =

(d) â&#x2C6;&#x2019;1

1 3 x 3

1 = â&#x2C6;&#x2019;1

Ď&#x20AC;/2

1 3 2 (1 â&#x2C6;&#x2019; (â&#x2C6;&#x2019;1)3 ) = = 2 3 3

Ď&#x20AC;/2

x2 dx;

Ď&#x20AC;/2

= sin(Ď&#x20AC;/2) â&#x2C6;&#x2019; sin(â&#x2C6;&#x2019;Ď&#x20AC;/2) = 1 + 1 = 2 = 2

cos xdx = sin x â&#x2C6;&#x2019;Ď&#x20AC;/2

â&#x2C6;&#x2019;Ď&#x20AC;/2

cos xdx 0

48. (a) cos 2x

x cos x

(b)

â&#x2C6;&#x161;

3x 3x2 + 1

(a) 0

54. F (x) = tanâ&#x2C6;&#x2019;1 x, F (x) = (a) 0

50. (a)

x = Ď&#x20AC;/4

1 â&#x2C6;&#x161; 1+ x

1 1 sin 2x â&#x2C6;&#x2019; , F (x) = cos 2x 2 2 (b)

ln x

52. |u|

3x2 + 1, F (x) = â&#x2C6;&#x161;

uh a

53. F (x) =

(b)

51. â&#x2C6;&#x2019;

â&#x2C6;&#x161; sin x

1 F (x) = sin 2t 2

49. (a)

46. The numerator is an odd function and the denominator is an even function, so the integrand is an odd function and the integral is zero. x 1 4 1 5 3 47. (a) x + 1 (b) F (x) = = x4 + x â&#x2C6;&#x2019; ; F (x) = x3 + 1 t +t 4 4 4 1

(b)

â&#x2C6;&#x161;

â&#x2C6;&#x161; (c) 6/ 13

1 1 + x2 (b) Ď&#x20AC;/3

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250

Chapter 6

xâ&#x2C6;&#x2019;3 = 0 when x = 3, which is a relative minimum, and hence the absolute x2 + 7 minimum, by the ďŹ rst derivative test.

55. (a) F (x) =

us uf

(b) increasing on [3, +â&#x2C6;&#x17E;), decreasing on (â&#x2C6;&#x2019;â&#x2C6;&#x17E;, 3]

7 + 6x â&#x2C6;&#x2019; x2 (7 â&#x2C6;&#x2019; x)(1 + x) = ; concave up on (â&#x2C6;&#x2019;1, 7), concave down on (â&#x2C6;&#x2019;â&#x2C6;&#x17E;, â&#x2C6;&#x2019;1) (x2 + 7)2 (x2 + 7)2 and on (7, +â&#x2C6;&#x17E;)

56.

3 2

t 20

-20 -10

57. (a) (0, +â&#x2C6;&#x17E;) because f is continuous there and 1 is in (0, +â&#x2C6;&#x17E;)

(b) at x = 1 because F (1) = 0

58. (a) (â&#x2C6;&#x2019;3, 3) because f is continuous there and 1 is in (â&#x2C6;&#x2019;3, 3) (b) at x = 1 because F (1) = 0 1 9

x1/2 dx = 2;

â&#x2C6;&#x161;

xâ&#x2C6;&#x2014; = 2, xâ&#x2C6;&#x2014; = 4

59. (a) fave =

2 1 2 1 (3x2 + 2x + 1) dx = (x3 + x2 + x) = 5; 3xâ&#x2C6;&#x2014;2 + 2xâ&#x2C6;&#x2014; + 1 = 5, 3 â&#x2C6;&#x2019;1 3 â&#x2C6;&#x2019;1 â&#x2C6;&#x161; â&#x2C6;&#x161; with solutions xâ&#x2C6;&#x2014; = â&#x2C6;&#x2019;(1/3)(1 Âą 13), but only xâ&#x2C6;&#x2014; = â&#x2C6;&#x2019;(1/3)(1 â&#x2C6;&#x2019; 13) lies in the interval [â&#x2C6;&#x2019;1, 2].

(b) fave =

Ď&#x20AC; 1 sin x dx = 0; sin xâ&#x2C6;&#x2014; = 0, xâ&#x2C6;&#x2014; = â&#x2C6;&#x2019;Ď&#x20AC;, 0, Ď&#x20AC; 2Ď&#x20AC; â&#x2C6;&#x2019;Ď&#x20AC; â&#x2C6;&#x161; 1 3 1 1 1 1 = dx = ; â&#x2C6;&#x2014; 2 = , xâ&#x2C6;&#x2014; = 3 2 1 x2 3 (x ) 3

60. (a) fave =

61.

â&#x2C6;&#x161;

2â&#x2030;¤

â&#x2C6;&#x161;

â&#x2C6;&#x161; 29, so 3 2 â&#x2030;¤

(b) fave

x3 + 2 â&#x2030;¤

â&#x2C6;&#x161; x3 + 2dx â&#x2030;¤ 3 29

Ď&#x20AC;

0â&#x2030;¤

62. Let f (x) = x sin x, f (0) = f (1) = 0, f (x) = sin x + x cos x = 0 when x = â&#x2C6;&#x2019; tan x, x â&#x2030;&#x2C6; 2.0288, so f has an absolute maximum at x â&#x2030;&#x2C6; 2.0288; f (2.0288) â&#x2030;&#x2C6; 1.8197, so 0 â&#x2030;¤ x sin x â&#x2030;¤ 1.82 and x sin xdx â&#x2030;¤ 1.82Ď&#x20AC; = 5.72

b b cF (x) a = cF (b) â&#x2C6;&#x2019; cF (a) = c[F (b) â&#x2C6;&#x2019; F (a)] = c F (x) a b (b) F (x) + G(x) a = [F (b) + G(b)] â&#x2C6;&#x2019; [F (a) + G(a)]

uh a

63. (a)

b F (x) â&#x2C6;&#x2019; G(x) a = [F (b) â&#x2C6;&#x2019; G(b)] â&#x2C6;&#x2019; [F (a) â&#x2C6;&#x2019; G(a)]

b b = [F (b) â&#x2C6;&#x2019; F (a)] â&#x2C6;&#x2019; [G(b) â&#x2C6;&#x2019; G(a)] = F (x) a â&#x2C6;&#x2019; G(x) a

(c)

b b = [F (b) â&#x2C6;&#x2019; F (a)] + [G(b) â&#x2C6;&#x2019; G(a)] = F (x) a + G(x) a

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Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Exercise Set 6.7

251

f (x) dx = F (b) â&#x2C6;&#x2019; F (a), i.e. a

f (x) dx = F (xâ&#x2C6;&#x2014; )(b â&#x2C6;&#x2019; a) = f (xâ&#x2C6;&#x2014; )(b â&#x2C6;&#x2019; a).

us uf

64. Let f be continuous on a closed interval [a, b] and let F be an antiderivative of f on [a, b]. By F (b) â&#x2C6;&#x2019; F (a) = F (xâ&#x2C6;&#x2014; ) for some xâ&#x2C6;&#x2014; in (a, b). By Theorem 6.6.1, Theorem 5.8.2, b â&#x2C6;&#x2019; a

n n Ď&#x20AC;k Ď&#x20AC; Ď&#x20AC;k Ď&#x20AC; Ď&#x20AC; 2 65. = f (xâ&#x2C6;&#x2014;k )â&#x2C6;&#x2020;x where f (x) = sec2 x, xâ&#x2C6;&#x2014;k = sec and â&#x2C6;&#x2020;x = for 0 â&#x2030;¤ x â&#x2030;¤ . 4n 4n 4n 4n 4 k=1 k=1 n n Ď&#x20AC;/4 Ď&#x20AC;/4 Ď&#x20AC; Ď&#x20AC;k lim sec2 f (xâ&#x2C6;&#x2014;k )â&#x2C6;&#x2020;x = sec2 x dx = tan x =1 Thus lim nâ&#x2020;&#x2019;+â&#x2C6;&#x17E; 4n nâ&#x2020;&#x2019;+â&#x2C6;&#x17E; 4n 0 0 n n n 1 k n 1 1 1 so , xâ&#x2C6;&#x2014; = , and â&#x2C6;&#x2020;x = = f (xâ&#x2C6;&#x2014;k )â&#x2C6;&#x2020;x where f (x) = = n n 1 + x2 k n2 + k 2 1 + k 2 /n2 n n2 + k 2 k=1 k=1 1 n n n 1 Ď&#x20AC; â&#x2C6;&#x2014; for 0 â&#x2030;¤ x â&#x2030;¤ 1. Thus lim = lim f (x )â&#x2C6;&#x2020;x = dx = . k 2 2 2 nâ&#x2020;&#x2019;+â&#x2C6;&#x17E; nâ&#x2020;&#x2019;+â&#x2C6;&#x17E; n +k 4 0 1+x k=1

k=1

66.

k=1

EXERCISE SET 6.7

1. (a) the increase in height in inches, during the ďŹ rst ten years (b) the change in the radius in centimeters, during the time interval t = 1 to t = 2 seconds (c) the change in the speed of sound in ft/s, during an increase in temperature from t = 32â&#x2014;Ś F to t = 100â&#x2014;Ś F (d) the displacement of the particle in cm, during the time interval t = t1 to t = t2 seconds 1

2. (a)

V (t)dt gal

(b) the change f (x1 ) â&#x2C6;&#x2019; f (x2 ) in the values of f over the interval

0 3

dist =

3. (a) displ = s(3) â&#x2C6;&#x2019; s(0) 2 3 3 2 3 2 2 = v(t)dt = (1 â&#x2C6;&#x2019; t)dt + (t â&#x2C6;&#x2019; 3)dt = (t â&#x2C6;&#x2019; t /2) + (t /2 â&#x2C6;&#x2019; 3t) = â&#x2C6;&#x2019;1/2; 0

(b) displ = s(3) â&#x2C6;&#x2019; s(0) 3 = v(t)dt =

dist =

m uh a 1

3 + (5t â&#x2C6;&#x2019; t2 ) = 3/2;

(2t â&#x2C6;&#x2019; 5)dt 5/2

5/2 3 2 + (5t â&#x2C6;&#x2019; t ) + (t â&#x2C6;&#x2019; 5t) 2

2 +t

(5 â&#x2C6;&#x2019; 2t)dt + 2

(5 â&#x2C6;&#x2019; 2t)dt = t2 /2

dt +

5/2

dt + 1 2

tdt +

+t 0

5/2

tdt +

= t /2

1 2 3 |v(t)|dt = (t â&#x2C6;&#x2019; t2 /2) + (t2 /2 â&#x2C6;&#x2019; t) â&#x2C6;&#x2019; (t2 /2 â&#x2C6;&#x2019; 3t) = 3/2

t 8

-1

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252

Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Chapter 6

5. (a) v(t) = 20 +

a(u)du; add areas of the small blocks to get 0

us uf

v(4) â&#x2030;&#x2C6; 20 + 1.4 + 3.0 + 4.7 + 6.2 = 35.3 m/s 6 (b) v(6) = v(4) + a(u)du â&#x2030;&#x2C6; 35.3 + 7.5 + 8.6 = 51.4 m/s 4

6. a > 0 and therefore (Theorem 6.5.6(a)) v > 0, so the particle is always speeding up for 0 < t < 10 (t3 â&#x2C6;&#x2019; 2t2 + 1)dt =

1 4 2 3 t â&#x2C6;&#x2019; t + t + C, 3 4

7. (a) s(t) =

2 1 1 4 2 3 (0) â&#x2C6;&#x2019; (0) + 0 + C = 1, C = 1, s(t) = t4 â&#x2C6;&#x2019; t3 + t + 1 3 4 3 4 (b) v(t) = 4 cos 2t dt = 2 sin 2t + C1 , v(0) = 2 sin 0 + C1 = â&#x2C6;&#x2019;1, C1 = â&#x2C6;&#x2019;1, v(t) = 2 sin 2t â&#x2C6;&#x2019; 1, s(t) = (2 sin 2t â&#x2C6;&#x2019; 1)dt = â&#x2C6;&#x2019; cos 2t â&#x2C6;&#x2019; t + C2 ,

s(0) =

s(0) = â&#x2C6;&#x2019; cos 0 â&#x2C6;&#x2019; 0 + C2 = â&#x2C6;&#x2019;3, C2 = â&#x2C6;&#x2019;2, s(t) = â&#x2C6;&#x2019; cos 2t â&#x2C6;&#x2019; t â&#x2C6;&#x2019; 2

(1 + sin t)dt = t â&#x2C6;&#x2019; cos t + C, s(0) = 0 â&#x2C6;&#x2019; cos 0 + C = â&#x2C6;&#x2019;3, C = â&#x2C6;&#x2019;2, s(t) = t â&#x2C6;&#x2019; cos t â&#x2C6;&#x2019; 2

8. (a) s(t) =

(t2 â&#x2C6;&#x2019; 3t + 1)dt =

(b) v(t) =

1 3 3 2 t â&#x2C6;&#x2019; t + t + C1 , 3 2

1 3 1 3 3 2 (0) â&#x2C6;&#x2019; (0) + 0 + C1 = 0, C1 = 0, v(t) = t3 â&#x2C6;&#x2019; t2 + t, 3 2 3 2 1 4 1 3 1 2 1 3 3 2 s(t) = t â&#x2C6;&#x2019; t + t dt = t â&#x2C6;&#x2019; t + t + C2 , 3 2 12 2 2 1 1 1 4 1 3 1 2 1 s(0) = (0)4 â&#x2C6;&#x2019; (0)3 + (0)2 + C2 = 0, C2 = 0, s(t) = t â&#x2C6;&#x2019; t + t 12 2 2 12 2 2 9. (a) s(t) = (2t â&#x2C6;&#x2019; 3)dt = t2 â&#x2C6;&#x2019; 3t + C, s(1) = (1)2 â&#x2C6;&#x2019; 3(1) + C = 5, C = 7, s(t) = t2 â&#x2C6;&#x2019; 3t + 7

v(0) =

(b) v(t) =

cos tdt = sin t + C1 , v(Ď&#x20AC;/2) = 2 = 1 + C1 , C1 = 1, v(t) = sin t + 1,

(sin t + 1)dt = â&#x2C6;&#x2019; cos t + t + C2 , s(Ď&#x20AC;/2) = 0 = Ď&#x20AC;/2 + C2 , C2 = â&#x2C6;&#x2019;Ď&#x20AC;/2,

s(t) =

s(t) = â&#x2C6;&#x2019; cos t + t â&#x2C6;&#x2019; Ď&#x20AC;/2

3 5/3 3 96 96 3 t + C, s(8) = 0 = 32 + C, C = â&#x2C6;&#x2019; , s(t) = t5/3 â&#x2C6;&#x2019; 5 5 5 5 5 â&#x2C6;&#x161; 2 3/2 2 13 2 3/2 13 tdt = t + C1 , v(4) = 1 = 8 + C1 , C1 = â&#x2C6;&#x2019; , v(t) = t â&#x2C6;&#x2019; , (b) v(t) = 3 3 3 3 3 4 5/2 13 13 4 44 2 3/2 13 t â&#x2C6;&#x2019; dt = t â&#x2C6;&#x2019; t + C2 , s(4) = â&#x2C6;&#x2019;5 = 32 â&#x2C6;&#x2019; 4 + C2 = â&#x2C6;&#x2019; + C2 , s(t) = 3 3 15 3 15 3 5 4 5/2 13 19 19 , s(t) = t â&#x2C6;&#x2019; t+ C2 = 5 15 3 5 t2/3 dt =

uh a

10. (a) s(t) =

11. (a) displacement = s(Ď&#x20AC;/2) â&#x2C6;&#x2019; s(0) = Ď&#x20AC;/2 distance = | sin t|dt = 1 m

Ď&#x20AC;/2

sin tdt = â&#x2C6;&#x2019; cos t 0

=1m 0

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Exercise Set 6.7

2Ď&#x20AC;

| cos t|dt = â&#x2C6;&#x2019;

distance =

Ď&#x20AC;/2 2Ď&#x20AC;

cos tdt +

Ď&#x20AC;/2

= â&#x2C6;&#x2019;1 m

cos tdt = sin t

Ď&#x20AC;/2 3Ď&#x20AC;/2

cos tdt = 3 m

Ď&#x20AC;/2

(b) displacement = s(2Ď&#x20AC;) â&#x2C6;&#x2019; s(Ď&#x20AC;/2) =

2Ď&#x20AC;

3Ď&#x20AC;/2

us uf

253

6 6 (2t â&#x2C6;&#x2019; 4)dt = (t2 â&#x2C6;&#x2019; 4t) = 12 m 12. (a) displacement = s(6) â&#x2C6;&#x2019; s(0) = 0 0 2 6 6 6 2 2 2 distance = (4 â&#x2C6;&#x2019; 2t)dt + (2t â&#x2C6;&#x2019; 4)dt = (4t â&#x2C6;&#x2019; t ) + (t â&#x2C6;&#x2019; 4t) = 20 m |2t â&#x2C6;&#x2019; 4|dt = 0

â&#x2C6;&#x2019;(t â&#x2C6;&#x2019; 3)dt +

|t â&#x2C6;&#x2019; 3|dt =

(b) displacement =

(t â&#x2C6;&#x2019; 3)dt = 13/2 m 3

|t â&#x2C6;&#x2019; 3|dt = 13/2 m

distance = 0

|v(t)|dt =

distance =

0 3

(b) displacement =

â&#x2C6;&#x161; â&#x2C6;&#x161; ( t â&#x2C6;&#x2019; 2)dt = 2 3 â&#x2C6;&#x2019; 6 m

|v(t)|dt = â&#x2C6;&#x2019;

distance =

v(t)dt = 11/4 m

â&#x2C6;&#x161; v(t)dt = 6 â&#x2C6;&#x2019; 2 3 m

â&#x2C6;&#x2019;v(t)dt +

v(t)dt +

v(t) = t3 â&#x2C6;&#x2019; 3t2 + 2t = t(t â&#x2C6;&#x2019; 1)(t â&#x2C6;&#x2019; 2) 3 displacement = (t3 â&#x2C6;&#x2019; 3t2 + 2t)dt = 9/4 m

13. (a)

1 1 ( â&#x2C6;&#x2019; 2 )dt = 1/3 m 2 t 1 â&#x2C6;&#x161;2 3 3 â&#x2C6;&#x161; |v(t)|dt = â&#x2C6;&#x2019; v(t)dt + â&#x2C6;&#x161; v(t)dt = 10/3 â&#x2C6;&#x2019; 2 2 m distance = 1

(b) displacement = 4

14. (a) displacement =

3tâ&#x2C6;&#x2019;1/2 dt = 6 m

v(t)dt = 6 m

v(t) = â&#x2C6;&#x2019;2t + 3 4 (â&#x2C6;&#x2019;2t + 3)dt = â&#x2C6;&#x2019;6 m displacement =

15.

|v(t)|dt =

distance =

| â&#x2C6;&#x2019; 2t + 3|dt =

distance = 1

3/2

(2t â&#x2C6;&#x2019; 3)dt = 13/2 m

(â&#x2C6;&#x2019;2t + 3)dt + 1

3/2

1 2 t â&#x2C6;&#x2019; 2t 2 5 1 2 t â&#x2C6;&#x2019; 2t dt = â&#x2C6;&#x2019;10/3 m displacement = 2 1 5 4 5 1 2 1 2 1 2 t â&#x2C6;&#x2019; 2t dt = distance = â&#x2C6;&#x2019; â&#x2C6;&#x2019; 2t dt + â&#x2C6;&#x2019; 2t dt = 17/3 m t t 2 2 2 1 1 4

uh a M

16.

v(t) =

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Chapter 6

2â&#x2C6;&#x161; 8 5t + 1 + 5 5 3 3 4 2â&#x2C6;&#x161; 8 8 3/2 dt = (5t + 1) + t = 204/25 m 5t + 1 + displacement = 5 5 75 5 0 0 3 3 distance = |v(t)|dt = v(t)dt = 204/25 m 0

18.

v(t) = â&#x2C6;&#x2019; cos t + 2 Ď&#x20AC;/2 â&#x2C6;&#x161; displacement = (â&#x2C6;&#x2019; cos t + 2)dt = (Ď&#x20AC; + 2 â&#x2C6;&#x2019; 2)/2 m Ď&#x20AC;/4

Ď&#x20AC;/2

| â&#x2C6;&#x2019; cos t + 2|dt =

distance =

(â&#x2C6;&#x2019; cos t + 2)dt = (Ď&#x20AC; +

â&#x2C6;&#x161;

2 â&#x2C6;&#x2019; 2)/2 m

Ď&#x20AC;/4

v(t) =

us uf

17.

Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

254

2 1 1 sin Ď&#x20AC;t dt = â&#x2C6;&#x2019; cos Ď&#x20AC;t + C 2 Ď&#x20AC; 2 2 2 2 1 s = 0 when t = 0 which gives C = so s = â&#x2C6;&#x2019; cos Ď&#x20AC;t + . 2 Ď&#x20AC; Ď&#x20AC; Ď&#x20AC; dv Ď&#x20AC; 1 a= = cos Ď&#x20AC;t. When t = 1 : s = 2/Ď&#x20AC;, v = 1, |v| = 1, a = 0. dt 2 2 3 3 (b) v = â&#x2C6;&#x2019;3 t dt = â&#x2C6;&#x2019; t2 + C1 , v = 0 when t = 0 which gives C1 = 0 so v = â&#x2C6;&#x2019; t2 2 2 3 1 1 t2 dt = â&#x2C6;&#x2019; t3 + C2 , s = 1 when t = 0 which gives C2 = 1 so s = â&#x2C6;&#x2019; t3 + 1. s=â&#x2C6;&#x2019; 2 2 2 When t = 1 : s = 1/2, v = â&#x2C6;&#x2019;3/2, |v| = 3/2, a = â&#x2C6;&#x2019;3.

19. (a) s =

20. (a) negative, because v is decreasing (b) speeding up when av > 0, so 2 < t < 5; slowing down when 1 < t < 2 (c) negative, because the area between the graph of v(t) and the t-axis appears to be greater where v < 0 compared to where v > 0

Ď&#x20AC;

3Ď&#x20AC;/2

sin xdx â&#x2C6;&#x2019;

(x2 â&#x2C6;&#x2019; 1)dx = 2/3 + 20/3 = 22/3

22. A = A1 + A2 =

(1 â&#x2C6;&#x2019; x2 )dx +

21. A = A1 + A2 =

sin xdx = 2 + 1 = 3

Ď&#x20AC;

1 â&#x2C6;&#x161; â&#x2C6;&#x161; 1 â&#x2C6;&#x2019; x + 1 dx + [ x + 1 â&#x2C6;&#x2019; 1] dx â&#x2C6;&#x2019;1 0 â&#x2C6;&#x161; â&#x2C6;&#x161; 0 1 2â&#x2C6;&#x2019;1 2 2 2 4 2 2 3/2 3/2 + =â&#x2C6;&#x2019; +1+ = x â&#x2C6;&#x2019; (x + 1) (x + 1) â&#x2C6;&#x2019; x â&#x2C6;&#x2019;1â&#x2C6;&#x2019; =4 3 3 3 3 3 3 0

23. A = A1 + A2 =

â&#x2C6;&#x2019;1

2 1 2 2 1 â&#x2C6;&#x2019; x2 x â&#x2C6;&#x2019;1 1 1 24. A = A1 + A2 = dx + dx = â&#x2C6;&#x2019; â&#x2C6;&#x2019; x + x + x2 x2 x x 1/2 1 1/2 1 1 1 = â&#x2C6;&#x2019;2 + 2 + + 2 + â&#x2C6;&#x2019; 2 = 1 2 2 1

uh a

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Exercise Set 6.7

(1 â&#x2C6;&#x2019; ex )dx +

26. A = A1 + A2 = 1/2

(ex â&#x2C6;&#x2019; 1)dx = 1/e + e â&#x2C6;&#x2019; 2 0

1â&#x2C6;&#x2019;x dx + x

xâ&#x2C6;&#x2019;1 dx = â&#x2C6;&#x2019; x

1 â&#x2C6;&#x2019; ln 2 + (1 â&#x2C6;&#x2019; ln 2) = 1/2 2

â&#x2C6;&#x2019;1

us uf

25. A = A1 + A2 =

255

and the displacement equals 4 cm if 4T 3/2 â&#x2C6;&#x2019; ln T = 11/2, T â&#x2030;&#x2C6; 1.272 s

27. By inspection the velocity is positive for t > 0, and during the ďŹ rst second the particle is at most 5/2 cm from the starting position. For T > 1 the displacement of the particle during the time interval [0, T ] is given by T T â&#x2C6;&#x161; T v(t) dt = 5/2 + (6 t â&#x2C6;&#x2019; 1/t) dt = 5/2 + (4t3/2 â&#x2C6;&#x2019; ln t) = â&#x2C6;&#x2019;3/2 + 4T 3/2 â&#x2C6;&#x2019; ln T ,

28. The displacement of the particle during the time interval [0, T ] is given by T v(t)dt = 3 tanâ&#x2C6;&#x2019;1 T â&#x2C6;&#x2019; 0.25T 2 . The particle is 2 cm from its starting position when 0

3 tanâ&#x2C6;&#x2019;1 T â&#x2C6;&#x2019; 0.25T 2 = 2 or when 3 tanâ&#x2C6;&#x2019;1 T â&#x2C6;&#x2019; 0.25T 2 = â&#x2C6;&#x2019;2; solve for T to get T = 0.90, 2.51, and 4.95 s.

150

20 3 20 3 t â&#x2C6;&#x2019; 50t2 + 50t + s0 , s(0) = 0 gives s0 = 0, so s(t) = t â&#x2C6;&#x2019; 50t2 + 50t, a(t) = 40t â&#x2C6;&#x2019; 100 3 3

29. s(t) =

-100

6 0

uh a

1200

2 30. v(t) = 2t2 â&#x2C6;&#x2019; 30t + v0 , v(0) = 3 = v0 , so v(t) = 2t2 â&#x2C6;&#x2019; 30t + 3, s(t) = t3 â&#x2C6;&#x2019; 15t2 + 3t + s0 , 3 2 3 2 s(0) = â&#x2C6;&#x2019;5 = s0 , so s(t) = t â&#x2C6;&#x2019; 15t + 3t â&#x2C6;&#x2019; 5 3

s(t)

25 0

-200

v(t)

-30

a(t)

-1200

500

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Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Chapter 6

y 3 2

31. (a) From the graph the velocity is at ﬁrst positive, but then turns negative, then positive again. The displacement, which is the cumulative area from x = 0 to x = 5, starts positive, turns negative, and then turns positive again.

us uf

256

–1

(b) displ = 5/2 − sin 5 + 5 cos 5 32. (a) If t0 < 1 then the area between the velocity curve and the t-axis, between t = 0 and t = t0 , will always be positive, so the displacement will be positive.

π2 + 4 2π 2

(b) displ =

−t

(b) s(t) = t/2 + (t + 1)e

0, −10,

-5

uh a

-10

0.2

-0.2

t2 t 1 ln(t + 0.1) − + − ln 10 4 20 200

t<4 t>4

0.1

-0.1

(b) v(t) =

25, t < 4 65 − 10t, t > 4

0.2 x

35. (a) a(t) =

(b) s(t) =

t2 1 − 2 200

0.4

34. (a) If t0 < 1 then the area between the velocity curve and the t-axis, between t = 0 and t = t0 , will always be negative, so the displacement will be negative.

0.8

33. (a) From the graph the velocity is positive, so the displacement is always increasing and is therefore positive.

0.4

20 2 4

t 8 10 12

-20 -40

25t, t<4 , so x(8) = 120, x(12) = −20 65t − 5t2 − 80, t > 4

(d) x(6.5) = 131.25

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0.6

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Exercise Set 6.7

v â&#x2C6;&#x2019; v0 ; from that and (8) a v â&#x2C6;&#x2019; v0 1 (v â&#x2C6;&#x2019; v0 )2 s â&#x2C6;&#x2019; s0 = v0 ; multiply through by a to get + a a 2 a2 1 1 1 2 a(s â&#x2C6;&#x2019; s0 ) = v0 (v â&#x2C6;&#x2019; v0 ) + (v â&#x2C6;&#x2019; v0 ) = (v â&#x2C6;&#x2019; v0 ) v0 + (v â&#x2C6;&#x2019; v0 ) = (v 2 â&#x2C6;&#x2019; v02 ). Thus 2 2 2

257

(c) From (9) v0 = v â&#x2C6;&#x2019; at; use this in (8) to get 1 1 s â&#x2C6;&#x2019; s0 = (v â&#x2C6;&#x2019; at)t + at2 = vt â&#x2C6;&#x2019; at2 2 2 This expression contains no v0 terms and so diďŹ&#x20AC;ers from (8).

(b) Put the last result of Part (a) into the ďŹ rst equation of Part (a) to obtain v â&#x2C6;&#x2019; v0 2(s â&#x2C6;&#x2019; s0 ) 2(s â&#x2C6;&#x2019; s0 ) t= = (v â&#x2C6;&#x2019; v0 ) 2 = . a v â&#x2C6;&#x2019; v02 v + v0

us uf

v 2 â&#x2C6;&#x2019; v02 . 2(s â&#x2C6;&#x2019; s0 )

36. (a) From (9) t =

37. (a) a = â&#x2C6;&#x2019;1 mi/h/s = â&#x2C6;&#x2019;22/15 ft/s2

(b) a = 30 km/h/min = 1/7200 km/s2

38. Take t = 0 when deceleration begins, then a = â&#x2C6;&#x2019;10 so v = â&#x2C6;&#x2019;10t + C1 , but v = 88 when t = 0 which gives C1 = 88 thus v = â&#x2C6;&#x2019;10t + 88, t â&#x2030;Ľ 0 (a) v = 45 mi/h = 66 ft/s, 66 = â&#x2C6;&#x2019;10t + 88, t = 2.2 s

(b) v = 0 (the car is stopped) when t = 8.8 s s = v dt = (â&#x2C6;&#x2019;10t + 88)dt = â&#x2C6;&#x2019;5t2 + 88t + C2 , and taking s = 0 when t = 0, C2 = 0 so

s = â&#x2C6;&#x2019;5t2 + 88t. At t = 8.8, s = 387.2. The car travels 387.2 ft before coming to a stop.

39. a = a0 ft/s2 , v = a0 t + v0 = a0 t + 132 ft/s, s = a0 t2 /2 + 132t + s0 = a0 t2 /2 + 132t ft; s = 200 ft 121 20 when v = 88 ft/s. Solve 88 = a0 t + 132 and 200 = a0 t2 /2 + 132t to get a0 = â&#x2C6;&#x2019; when t = , 5 11 121 t + 132. so s = â&#x2C6;&#x2019;12.1t2 + 132t, v = â&#x2C6;&#x2019; 5 121 242 70 ft/s2 (a) a0 = â&#x2C6;&#x2019; (b) v = 55 mi/h = ft/s when t = s 5 3 33 60 s (c) v = 0 when t = 11

40. dv/dt = 3, v = 3t + C1 , but v = v0 when t = 0 so C1 = v0 , v = 3t + v0 . From ds/dt = v = 3t + v0 we get s = 3t2 /2 + v0 t + C2 and, with s = 0 when t = 0, C2 = 0 so s = 3t2 /2 + v0 t. s = 40 when t = 4 thus 40 = 3(4)2 /2 + v0 (4), v0 = 4 m/s

41. Suppose s = s0 = 0, v = v0 = 0 at t = t0 = 0; s = s1 = 120, v = v1 at t = t1 ; and s = s2 , v = v2 = 12 at t = t2 . From Exercise 36(a), v12 â&#x2C6;&#x2019; v02 , v 2 = 2as1 = 5.2(120) = 624. Applying the formula again, 2(s1 â&#x2C6;&#x2019; s0 ) 1 v 2 â&#x2C6;&#x2019; v12 , v 2 = v12 â&#x2C6;&#x2019; 3(s2 â&#x2C6;&#x2019; s1 ), so â&#x2C6;&#x2019;1.5 = a = 2 2(s2 â&#x2C6;&#x2019; s1 ) 2

uh a

2.6 = a =

s2 = s1 â&#x2C6;&#x2019; (v22 â&#x2C6;&#x2019; v12 )/3 = 120 â&#x2C6;&#x2019; (144 â&#x2C6;&#x2019; 624)/3 = 280 m.

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Chapter 6

4t, t < 2 t<2 and, , so, with v0 = 0, v(t) = 8, t > 2 t>2 2t2 , t < 2 since s0 = 0, s(t) = s = 100 when 8t − 8 = 100, t = 108/8 = 13.5 s 8t − 8, t > 2

4, 0,

us uf

42. a(t) =

258

43. The truck’s velocity is vT = 50 and its position is sT = 50t+5000. The car’s acceleration is aC = 2, so vC = 2t, sC = t2 (initial position and initial velocity of the car are both zero). sT = sC when 50t + 5000 = t2 , t2 − 50t − 5000 = (t + 50)(t − 100) = 0, t = 100 s and sC = sT = t2 = 10, 000 ft.

44. Let t = 0 correspond to the time when the leader is 100 m from the ﬁnish line; let s = 0 correspond to the ﬁnish line. Then vC = 12, sC = 12t − 115; aL = 0.5 for t > 0, vL√= 0.5t + 8, sL = 0.25t2 + 8t − 100. sC = 0 at t = 115/12 ≈ 9.58 s, and sL = 0 at t = −16 + 4 41 ≈ 9.61, so the challenger wins. 45. s = 0 and v = 112 when t = 0 so v(t) = −32t + 112, s(t) = −16t2 + 112t

(a) v(3) = 16 ft/s, v(5) = −48 ft/s (b) v = 0 when the projectile is at its maximum height so −32t + 112 = 0, t = 7/2 s, s(7/2) = −16(7/2)2 + 112(7/2) = 196 ft. (c) s = 0 when it reaches the ground so −16t2 + 112t = 0, −16t(t − 7) = 0, t = 0, 7 of which t = 7 is when it is at ground level on its way down. v(7) = −112, |v| = 112 ft/s.

46. s = 112 when t = 0 so s(t) = −16t2 + v0 t + 112. But s = 0 when t = 2 thus −16(2)2 + v0 (2) + 112 = 0, v0 = −24 ft/s. 47. (a) s(t) = 0 when it hits the ground, s(t) = −16t2 + 16t = −16t(t − 1) = 0 when t = 1 s.

(b) The projectile moves upward until it gets to its highest point where v(t) = 0, v(t) = −32t + 16 = 0 when t = 1/2 s. √

555/4 s

48. (a) s(t) = 0 when the rock hits the ground, s(t) = −16t2 + 555 = 0 when t = √ √ √ (b) v(t) = −32t, v( 555/4) = −8 555, the speed at impact is 8 555 ft/s

49. (a) s(t) = 0 when the package hits the ground, √ s(t) = −16t2 + 20t + 200 = 0 when t = (5 + 5 33)/8 s √ √ √ (b) v(t) = −32t + 20, v[(5 + 5 33)/8] = −20 33, the speed at impact is 20 33 ft/s 50. (a) s(t) = 0 when the stone hits the ground, s(t) = −16t2 − 96t + 112 = −16(t2 + 6t − 7) = −16(t + 7)(t − 1) = 0 when t = 1 s

(b) v(t) = −32t − 96, v(1) = −128, the speed at impact is 128 ft/s 51. s(t) = −4.9t2 + 49t + 150 and v(t) = −9.8t + 49

(a) the projectile reaches its maximum height when v(t) = 0, −9.8t + 49 = 0, t = 5 s (b) s(5) = −4.9(5)2 + 49(5) + 150 = 272.5 m

uh a

(c) the projectile reaches its starting point when s(t) = 150, −4.9t2 + 49t + 150 = 150, −4.9t(t − 10) = 0, t = 10 s (d) v(10) = −9.8(10) + 49 = −49 m/s

(e) s(t) = 0 when the projectile hits the ground, −4.9t2 + 49t + 150 = 0 when (use the quadratic formula) t ≈ 12.46 s (f )

v(12.46) = −9.8(12.46) + 49 ≈ −73.1, the speed at impact is about 73.1 m/s

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Exercise Set 6.7

259

52. take s = 0 at the water level and let h be the height of the bridge, then s = h and v = 0 when t = 0 so s(t) = â&#x2C6;&#x2019;16t2 + h

(a) s = 0 when t = 4 thus â&#x2C6;&#x2019;16(4)2 + h = 0, h = 256 ft

us uf

2 (b) First, â&#x2C6;&#x161; ďŹ nd how long it takes for the stone to hit the water (ďŹ nd t for s = 0) : â&#x2C6;&#x2019;16t + h = 0, t = h/4. Next, ďŹ nd how long it takes the sound to travel to the bridge: this time is h/1080 because the speed is constant at 1080 ft/s. Finally, use the fact that the total of these two â&#x2C6;&#x161; â&#x2C6;&#x161; â&#x2C6;&#x161; h h + = 4, h + 270 h = 4320, h + 270 h â&#x2C6;&#x2019; 4320 = 0, and by times must be 4 s: 1080 4 â&#x2C6;&#x161; â&#x2C6;&#x2019;270 Âą (270)2 + 4(4320) , reject the negative value to get the quadratic formula h = 2 â&#x2C6;&#x161; h â&#x2030;&#x2C6; 15.15, h â&#x2030;&#x2C6; 229.5 ft.

53. g = 9.8/6 = 4.9/3 m/s2 , so v = â&#x2C6;&#x2019;(4.9/3)t, s = â&#x2C6;&#x2019;(4.9/6)t2 + 5, s = 0 when t = 30/4.9 and v = â&#x2C6;&#x2019;(4.9/3) 30/4.9 â&#x2030;&#x2C6; â&#x2C6;&#x2019;4.04, so the speed of the module upon landing is 4.04 m/s

59. fave =

1 eâ&#x2C6;&#x2019;1

Ď&#x20AC;

3 3x dx = x2 4

cos x dx = 0 e

56. fave

1 x dx = x3 9 â&#x2C6;&#x2019;1

Ď&#x20AC; 1 sin x = 0 Ď&#x20AC; 0

ln 5

ex dx =

â&#x2C6;&#x2019;1

1 5 â&#x2C6;&#x2019; eâ&#x2C6;&#x2019;1 (5 â&#x2C6;&#x2019; eâ&#x2C6;&#x2019;1 ) = ln 5 + 1 1 + ln 5

x2 dx = 4/3

â&#x2C6;&#x161; (b) (xâ&#x2C6;&#x2014; )2 = 4/3,â&#x2C6;&#x161;xâ&#x2C6;&#x2014; = Âą2/ 3, but only 2/ 3 is in [0, 2]

1 = 2â&#x2C6;&#x2019;0 y

(c)

uh a

62. (a) fave

x 2 3

1 = 4â&#x2C6;&#x2019;0

2x dx = 4

1 1 1 dx = (ln e â&#x2C6;&#x2019; ln 1) = x eâ&#x2C6;&#x2019;1 eâ&#x2C6;&#x2019;1

1 60. fave = ln 5 â&#x2C6;&#x2019; (â&#x2C6;&#x2019;1) 61. (a) fave

=6 1

1 = 2 â&#x2C6;&#x2019; (â&#x2C6;&#x2019;1)

Ď&#x20AC; 1 sin x dx = â&#x2C6;&#x2019; cos x = 2/Ď&#x20AC; Ď&#x20AC; 0

Ď&#x20AC;

1 Ď&#x20AC;â&#x2C6;&#x2019;0

58. fave =

57. fave

1 = Ď&#x20AC;â&#x2C6;&#x2019;0

55. fave

1 = 3â&#x2C6;&#x2019;1

54. s(t) = â&#x2C6;&#x2019; 12 gt2 + v0 t; s = 1000 when v = 0, so 0 = v = â&#x2C6;&#x2019;gt + v0 , t = v0 /g, â&#x2C6;&#x161; 1000 = s(v0 /g) = â&#x2C6;&#x2019; 12 g(v0 /g)2 + v0 (v0 /g) = 12 v02 /g, so v02 = 2000g, v0 = 2000g. â&#x2C6;&#x161; The initial velocity on the Earth would have to be 6 times faster than that on the Moon.

(b) 2xâ&#x2C6;&#x2014; = 4, xâ&#x2C6;&#x2014; = 2

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=1 â&#x2C6;&#x2019;1

260

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Chapter 6

(c)

us uf

63. (a) vave =

(b) aave

(3t3 + 2)dt = 1

263 1 789 = 3 4 4

s(4) â&#x2C6;&#x2019; s(1) 100 â&#x2C6;&#x2019; 7 = = = 31 4â&#x2C6;&#x2019;1 3 1 = 5â&#x2C6;&#x2019;0

(t + 1)dt = 7/2 0

â&#x2C6;&#x161;

v(Ď&#x20AC;/4) â&#x2C6;&#x2019; v(0) = = Ď&#x20AC;/4 â&#x2C6;&#x2019; 0

64. (a) aave

1 4â&#x2C6;&#x2019;1

â&#x2C6;&#x161; 2/2 â&#x2C6;&#x2019; 1 = (2 2 â&#x2C6;&#x2019; 4)/Ď&#x20AC; Ď&#x20AC;/4

(b) vave

65. time to ďŹ ll tank = (volume of tank)/(rate of ďŹ lling) = [Ď&#x20AC;(3)2 5]/(1) = 45Ď&#x20AC;, weight of water in tank at time t = (62.4) (rate of ďŹ lling)(time) = 62.4t, 45Ď&#x20AC; 1 weightave = 62.4t dt = 1404Ď&#x20AC; lb 45Ď&#x20AC; 0

66. (a) If x is the distance from the cooler end, then the temperature is T (x) = (15 + 1.5x)â&#x2014;Ś C, and 10 1 Tave = (15 + 1.5x)dx = 22.5â&#x2014;Ś C 10 â&#x2C6;&#x2019; 0 0

(b) By the Mean-Value Theorem for Integrals there exists xâ&#x2C6;&#x2014; in [0, 10] such that 10 1 â&#x2C6;&#x2014; f (x ) = (15 + 1.5x)dx = 22.5, 15 + 1.5xâ&#x2C6;&#x2014; = 22.5, xâ&#x2C6;&#x2014; = 5 10 â&#x2C6;&#x2019; 0 0

67. (a) amount of water = (rate of ďŹ&#x201A;ow)(time) = 4t gal, total amount = 4(30) = 120 gal 60 (b) amount of water = (4 + t/10)dt = 420 gal 0 120 â&#x2C6;&#x161; â&#x2C6;&#x161; (10 + t)dt = 1200 + 160 30 â&#x2030;&#x2C6; 2076.36 gal (c) amount of water = 0

68. (a) The maximum value of R occurs at 4:30 P.M. when t = 0. 60 (b) 100(1 â&#x2C6;&#x2019; 0.0001t2 )dt = 5280 cars

69. (a)

uh a

f (x)dx â&#x2C6;&#x2019; a

because fave (b â&#x2C6;&#x2019; a) =

[f (x) â&#x2C6;&#x2019; fave ] dx =

f (x)dx â&#x2C6;&#x2019; fave (b â&#x2C6;&#x2019; a) = 0

fave dx = a

f (x)dx a

b b [f (x) â&#x2C6;&#x2019; c]dx = 0 then f (x)dx â&#x2C6;&#x2019; c(b â&#x2C6;&#x2019; a) = 0 so (b) no, because if a b a 1 c= f (x)dx = fave is the only value bâ&#x2C6;&#x2019;a a

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Exercise Set 6.8

261

EXERCISE SET 6.8

2. (a)

1 2

u1/2 du

(c)

1 Ï&#x20AC;

sin u du

â&#x2C6;&#x2019;3

(d)

eu du Ï&#x20AC;/3

â&#x2C6;&#x161;

u du 1/2

â&#x2C6;&#x161;

(b)

1 4

u3 du = 2

1 2

1 5 u 10 1 4 u 16

= 80, or 2

(u â&#x2C6;&#x2019; 1)u

du =

2 2 (1 + x)5/2 â&#x2C6;&#x2019; (1 + x)3/2 3 5

3/2

cos u du =

1 1 14. u = x â&#x2C6;&#x2019; , 4 4 4

â&#x2C6;&#x2019;3

= 19 1

9 = 1192/15, 1

2 8 )du = u5/2 â&#x2C6;&#x2019; u3/2 5 3

2 sin u 3

1 du = â&#x2C6;&#x2019; 2 4u

4 = â&#x2C6;&#x2019;506/15 9

Ï&#x20AC;/2 â&#x2C6;&#x161; â&#x2C6;&#x161; = 8 â&#x2C6;&#x2019; 4 2, or â&#x2C6;&#x2019; 8 cos(x/2) =8â&#x2C6;&#x2019;4 2 0

Ï&#x20AC;/2 = 2/3, or 0

3 6

Ï&#x20AC;/6 2 = 2/3 sin 3x 3 0

1 1 = â&#x2C6;&#x2019;1/48, or â&#x2C6;&#x2019; 4 (x2 + 2)2 Ï&#x20AC;/4

Ï&#x20AC;/4 2

sec u du = 4 tan u â&#x2C6;&#x2019;Ï&#x20AC;/4

Ï&#x20AC;/2

â&#x2C6;&#x2019; 4u

Ï&#x20AC;/4

1 13. u = x + 2, 2

= 10 â&#x2C6;&#x2019;1

= â&#x2C6;&#x2019;506/15

â&#x2C6;&#x2019;5

sin u du = â&#x2C6;&#x2019;8 cos u

2 2 )du = u5/2 â&#x2C6;&#x2019; u3/2 5 3

1/2

Ï&#x20AC;/4

11. u = x/2, 8

â&#x2C6;&#x2019;u

1/2

du =

2 8 (4 â&#x2C6;&#x2019; x)5/2 â&#x2C6;&#x2019; (4 â&#x2C6;&#x2019; x)3/2 5 3

= 80

= 1192/15

1/2

= 121/5

(u â&#x2C6;&#x2019; 4)u

10. u = 4 â&#x2C6;&#x2019; x,

uh a

1 = 19, or â&#x2C6;&#x2019; (4 â&#x2C6;&#x2019; 3x)9 27

â&#x2C6;&#x2019;2

1/2

1 (4x â&#x2C6;&#x2019; 2)4 16

1 = 10, or â&#x2C6;&#x2019; (1 â&#x2C6;&#x2019; 2x)4 8

1 u du = â&#x2C6;&#x2019; u9 27

9. u = 1 + x,

1 u3 du = â&#x2C6;&#x2019; u4 8

â&#x2C6;&#x2019;2

1 (2x + 1)5 10

= 121/5, or

du 1 â&#x2C6;&#x2019; u2

u4 du =

1 8. u = 4 â&#x2C6;&#x2019; 3x, â&#x2C6;&#x2019; 3

7. u = 1 â&#x2C6;&#x2019; 2x, â&#x2C6;&#x2019;

2 3

(u â&#x2C6;&#x2019; 3)u1/2 du 3

u du

1 2

6. u = 4x â&#x2C6;&#x2019; 2,

12. u = 3x,

5. u = 2x + 1,

1 â&#x2C6;&#x161; du u

Ï&#x20AC;/4

(b)

â&#x2C6;&#x2019;1

4. (a)

5/2

1 2

(u + 5)u20 du â&#x2C6;&#x2019;3

3/2

(d)

â&#x2C6;&#x2019;Ï&#x20AC;

(b)

u2 du 1 2

Ï&#x20AC;

u8 du

= 8, or 4 tan

â&#x2C6;&#x2019;Ï&#x20AC;/4

â&#x2C6;&#x2019;1 â&#x2C6;&#x2019;2

= â&#x2C6;&#x2019;1/48

1 1 xâ&#x2C6;&#x2019; 4 4

â&#x2C6;&#x2019;

(b)

us uf

u7 du

1. (a)

1+Ï&#x20AC; =8 1â&#x2C6;&#x2019;Ï&#x20AC;

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262

Chapter 6

13 1 +4= when x = â&#x2C6;&#x2019; ln 3, 3 3 7 7 1 ln 3 du = ln u u = e + 4 = 3 + 4 = 7 when x = ln 3, = ln(7) â&#x2C6;&#x2019; ln(13/3) = ln(21/13) 13/3 u 13/3

â&#x2C6;&#x161; 3/2

â&#x2C6;&#x161; 1/2

25 â&#x2C6;&#x2019;

21. â&#x2C6;&#x2019;

1 2

â&#x2C6;&#x2019;6

3 cos 2x dx =

(x + 5)â&#x2C6;&#x2019;2 dx = â&#x2C6;&#x2019;(x + 5)â&#x2C6;&#x2019;1

dx 1 =â&#x2C6;&#x2019; (3x + 1)2 3(3x + 1)

1/6

â&#x2C6;&#x161;

27. A = 0

1 1 dx = 3 1 â&#x2C6;&#x2019; 9x2

fave

16 â&#x2C6;&#x2019;

us uf

1 1 2 Ï&#x20AC;(4) = 2Ï&#x20AC; du = 2 4

1 1 Â· [Ï&#x20AC;(1)2 ] = Ï&#x20AC;/8 2 4

1 = 0

1/2

â&#x2C6;&#x161; 0

1 4

1 1 du = sinâ&#x2C6;&#x2019;1 u 3 1 â&#x2C6;&#x2019; u2

1/2 = Ï&#x20AC;/18 0

Ï&#x20AC;/2

sin y dy = â&#x2C6;&#x2019; cos y 0

=1 0

2 1 x 1 1 1 dx = â&#x2C6;&#x2019; = (5x2 + 1)2 2 10 5x2 + 1 21

uh a M

30.

1 2â&#x2C6;&#x2019;0

1 2

1 1 1 + = 12 8 24

Ï&#x20AC;/2

28. x = sin y, A =

29.

=â&#x2C6;&#x2019;

1/2

Ï&#x20AC;/8 â&#x2C6;&#x161; 3 sin 2x = 3 2/4 2 0

26.

1 â&#x2C6;&#x2019; u2 du =

Ï&#x20AC; Ï&#x20AC; 3 1 Ï&#x20AC; + sinâ&#x2C6;&#x2019;1 = â&#x2C6;&#x2019; + = â&#x2C6;&#x2019; 3 6 6 2 2

1 1 1 cos Ï&#x20AC;x = â&#x2C6;&#x2019; (â&#x2C6;&#x2019;1 â&#x2C6;&#x2019; 1) = 2/Ï&#x20AC; Ï&#x20AC; Ï&#x20AC; 0

0 7

Ï&#x20AC;/8

24. A =

1 2

â&#x2C6;&#x161; = â&#x2C6;&#x2019; sinâ&#x2C6;&#x2019;1

20.

36 â&#x2C6;&#x2019; u2 du = Ï&#x20AC;(6)2 /2 = 18Ï&#x20AC;

sin Ï&#x20AC;xdx = â&#x2C6;&#x2019;

25.

â&#x2C6;&#x161;3/2

25 1 1 2 Ï&#x20AC;(5) = Ï&#x20AC; du = 3 4 12

1 â&#x2C6;&#x2019; u2 du =

22.

23.

1 du = â&#x2C6;&#x2019; sinâ&#x2C6;&#x2019;1 u 1 â&#x2C6;&#x2019; u2

3 â&#x2C6;&#x2019; tanâ&#x2C6;&#x2019;1 1) = 2(Ï&#x20AC;/3 â&#x2C6;&#x2019; Ï&#x20AC;/4) = Ï&#x20AC;/6

19.

1 3

â&#x2C6;&#x161;

18. u = eâ&#x2C6;&#x2019;x , â&#x2C6;&#x2019;

= 2(tanâ&#x2C6;&#x2019;1

1 du = 2 tanâ&#x2C6;&#x2019;1 u u2 + 1

x, 2

â&#x2C6;&#x161;3

â&#x2C6;&#x161;

17. u =

â&#x2C6;&#x161; 3

16. u = 3 â&#x2C6;&#x2019; 4ex , du = â&#x2C6;&#x2019;4ex dx, u = â&#x2C6;&#x2019;1 when x = 0, u = â&#x2C6;&#x2019;17 when x = ln 5 â&#x2C6;&#x2019;17 1 â&#x2C6;&#x2019;17 1 â&#x2C6;&#x2019; u du = â&#x2C6;&#x2019; u2 = â&#x2C6;&#x2019;36 8 4 â&#x2C6;&#x2019;1 â&#x2C6;&#x2019;1

1 = 1/4 â&#x2C6;&#x2019; (â&#x2C6;&#x2019;1/4)

1/4 2 4 sec Ï&#x20AC;xdx = tan Ï&#x20AC;x = Ï&#x20AC; Ï&#x20AC; â&#x2C6;&#x2019;1/4

15. u = ex + 4, du = ex dx, u = eâ&#x2C6;&#x2019; ln 3 + 4 =

1/4

â&#x2C6;&#x2019;1/4

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Exercise Set 6.8

â&#x2C6;&#x161; 1/ 3

2 (3x + 1)1/2 3 2 3 (x + 9)1/2 3

du 2 tanâ&#x2C6;&#x2019;1 u = 1 + u2 ln 3

39.

1 â&#x2C6;&#x2019;1

â&#x2C6;&#x161; 2 â&#x2C6;&#x161; = ( 10 â&#x2C6;&#x2019; 2 2) 3

â&#x2C6;&#x2019;1/2

28 du = u

2 (tan x)3/2 3

40.

1 sec u du = tan u 3

1 3

â&#x2C6;&#x2019;1

u2 du =

= 38/15 1

= 1/10 â&#x2C6;&#x2019;1

â&#x2C6;&#x161; â&#x2C6;&#x161; 12 = 2( 7 â&#x2C6;&#x2019; 3)

Ď&#x20AC;/4

= 2/3 0

â&#x2C6;&#x161;Ď&#x20AC; 5 sin(x2 ) =0 2 0

41.

Ď&#x20AC;/3

â&#x2C6;&#x161; = ( 3 â&#x2C6;&#x2019; 1)/3

Ď&#x20AC;/4

44. u = sin 3Î¸,

â&#x2C6;&#x161;

= â&#x2C6;&#x2019;4 Ď&#x20AC;

Ď&#x20AC;/4

28 â&#x2C6;&#x2019;

= 1/2

2Ď&#x20AC;

Ď&#x20AC;/3

â&#x2C6;&#x161;

sin u du = â&#x2C6;&#x2019;2 cos u

Ď&#x20AC;

1 43. u = 3Î¸, 3

1/2 12

2Ď&#x20AC;

x, 2

1 3 (t + 1)20 10

36.

1 1 dx = â&#x2C6;&#x2019; (x â&#x2C6;&#x2019; 3)2 xâ&#x2C6;&#x2019;3

â&#x2C6;&#x161;

2 (5x â&#x2C6;&#x2019; 1)3/2 15

34.

Ď&#x20AC;/4 1 2 sin x =0 2 â&#x2C6;&#x2019;3Ď&#x20AC;/4

42. u =

â&#x2C6;&#x161; 1/ 3

Ď&#x20AC; 2 Ď&#x20AC; Ď&#x20AC; â&#x2C6;&#x2019; = ln 3 4 6 6 ln 3

= 2/3

1 37. u = x + 4x + 7, 2

38.

2 = ln 3

1 (1 â&#x2C6;&#x2019; eâ&#x2C6;&#x2019;8 ) 8

us uf

1 dx = â&#x2C6;&#x2019; eâ&#x2C6;&#x2019;2x 8

35.

â&#x2C6;&#x2019;2x

33.

32. fave

31. fave

1 = 4

1 3 u 9

â&#x2C6;&#x2019;1

= â&#x2C6;&#x2019;1/9

1 1 45. u = 4 â&#x2C6;&#x2019; 3y, y = (4 â&#x2C6;&#x2019; u), dy = â&#x2C6;&#x2019; du 3 3 1 4 1 16 â&#x2C6;&#x2019; 8u + u2 1 â&#x2C6;&#x2019; du = (16uâ&#x2C6;&#x2019;1/2 â&#x2C6;&#x2019; 8u1/2 + u3/2 )du 27 4 27 1 u1/2 4 1 16 3/2 2 5/2 1/2 32u â&#x2C6;&#x2019; u + u = = 106/405 3 5 27 1

46. u = 5 + x,

uh a

uâ&#x2C6;&#x2019;5 â&#x2C6;&#x161; du = u

1/2

e 47. ln(x + e) = ln(2e) â&#x2C6;&#x2019; ln e = ln 2 0

49. u =

â&#x2C6;&#x161;

3x2 ,

1 â&#x2C6;&#x161;

2 3

â&#x2C6;&#x161; 3

â&#x2C6;&#x161; 0

263

â&#x2C6;&#x2019; 5u

â&#x2C6;&#x2019;1/2

2 )du = u3/2 â&#x2C6;&#x2019; 10u1/2 3 2 1 â&#x2C6;&#x2019; eâ&#x2C6;&#x2019;x 2

48.

1 u 1 du = â&#x2C6;&#x161; sinâ&#x2C6;&#x2019;1 2 2 3 4 â&#x2C6;&#x2019; u2

â&#x2C6;&#x161;3 0

â&#x2C6;&#x161;2

9 = 8/3 4

= (eâ&#x2C6;&#x2019;1 â&#x2C6;&#x2019; eâ&#x2C6;&#x2019;2 )/2

Ď&#x20AC; 1 Ď&#x20AC; = â&#x2C6;&#x161; = â&#x2C6;&#x161; 2 3 3 6 3

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Chapter 6

x, 2 1

1 3

51. u = 3x,

1 u 1 du = tanâ&#x2C6;&#x2019;1 4 + u2 6 2

1 2

52. u = x2 ,

â&#x2C6;&#x161; 2 3

â&#x2C6;&#x161;2 = 2(Ï&#x20AC;/4 â&#x2C6;&#x2019; Ï&#x20AC;/6) = Ï&#x20AC;/6 1

2â&#x2C6;&#x161;3 = 0

u 1 1 du = â&#x2C6;&#x161; tanâ&#x2C6;&#x2019;1 â&#x2C6;&#x161; 3 + u2 3 2 3

Ï&#x20AC;/6 2

sin x(1 â&#x2C6;&#x2019; sin x) cos x dx =

53. (b)

1 u â&#x2C6;&#x161; du = 2 sinâ&#x2C6;&#x2019;1 2 2 4â&#x2C6;&#x2019;u

3 1

us uf

â&#x2C6;&#x161; 2

Ï&#x20AC; 1Ï&#x20AC; = 63 18

1 Ï&#x20AC; = â&#x2C6;&#x161; (Ï&#x20AC;/3 â&#x2C6;&#x2019; Ï&#x20AC;/6) = â&#x2C6;&#x161; 2 3 12 3

Ï&#x20AC;/6 1 1 1 1 23 7 5 sin x â&#x2C6;&#x2019; sin x â&#x2C6;&#x2019; = = 7 4480 5 160 896

50. u =

â&#x2C6;&#x161;

Ï&#x20AC;/4 Ï&#x20AC;/4 1 3 (sec2 x â&#x2C6;&#x2019; 1) dx tan x(sec x â&#x2C6;&#x2019; 1) dx = tan x â&#x2C6;&#x2019; 3 â&#x2C6;&#x2019;Ï&#x20AC;/4 â&#x2C6;&#x2019;Ï&#x20AC;/4 â&#x2C6;&#x2019;Ï&#x20AC;/4 Ï&#x20AC;/4 2 2 Ï&#x20AC; 4 Ï&#x20AC; = + (â&#x2C6;&#x2019; tan x + x) = â&#x2C6;&#x2019;2+ =â&#x2C6;&#x2019; + 3 2 3 2 3

Ï&#x20AC;/4

54. (b)

264

f (u)du = 5/3 1 0

(b) u = 3x, 0

1 m

m n

(1 â&#x2C6;&#x2019; u) u du =

Ï&#x20AC;/2 n

57. sin x = cos(Ï&#x20AC;/2 â&#x2C6;&#x2019; x), Ï&#x20AC;/2 Ï&#x20AC;/2 n n sin x dx = cos (Ï&#x20AC;/2 â&#x2C6;&#x2019; x)dx = â&#x2C6;&#x2019; 0

(1 â&#x2C6;&#x2019; u)un du =

xn (1 â&#x2C6;&#x2019; x)m dx 0

cosn u du

(u = Ï&#x20AC;/2 â&#x2C6;&#x2019; x)

Ï&#x20AC;/2

(by replacing u by x)

58. u = 1 â&#x2C6;&#x2019; x, â&#x2C6;&#x2019;

f (u)du = 5/3

cosn x dx

cos u du = 0

u (1 â&#x2C6;&#x2019; u) du =

x (1 â&#x2C6;&#x2019; x) dx = â&#x2C6;&#x2019;

56. u = 1 â&#x2C6;&#x2019; x,

f (u)du = â&#x2C6;&#x2019;1/2

1 3

f (u)du = â&#x2C6;&#x2019;1/2

1 3

55. (a) u = 3x + 1,

â&#x2C6;&#x2019;Ï&#x20AC;/4

(1 â&#x2C6;&#x2019; u)un du = 0

(un â&#x2C6;&#x2019; un+1 )du = 0

1 1 â&#x2C6;&#x2019; n+1 n+2

1 = (n + 1)(n + 2)

e1.528t dt = 524.959e1.528t + C; y(0) = 750 = 524.959 + C, C = 225.041,

59. y(t) = (802.137)

uh a

y(t) = 524.959e1.528t + 225.041, y(12) = 48, 233, 500, 000 60.

275000 10 â&#x2C6;&#x2019; 0

Vave =

61. s(t) =

eâ&#x2C6;&#x2019;0.17t dt = â&#x2C6;&#x2019;161764.7059eâ&#x2C6;&#x2019;0.17t

10 = $132, 212.96 0

(25 + 10eâ&#x2C6;&#x2019;0.05t )dt = 25t â&#x2C6;&#x2019; 200eâ&#x2C6;&#x2019;0.05t + C

â&#x2C6;&#x161; (a) s(10) â&#x2C6;&#x2019; s(0) = 250 â&#x2C6;&#x2019; 200(eâ&#x2C6;&#x2019;0.5 â&#x2C6;&#x2019; 1) = 450 â&#x2C6;&#x2019; 200/ e â&#x2030;&#x2C6; 328.69 ft (b) yes; without it the distance would have been 250 ft

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Exercise Set 6.8

1 e dx = 3, e2x 2

1 1 = 3, (e2k â&#x2C6;&#x2019; 1) = 3, e2k = 7, k = ln 7 2 2

62. 0

63. The area is given by

â&#x2C6;&#x161; â&#x2C6;&#x161; 1/(1 + kx2 )dx = (1/ k) tanâ&#x2C6;&#x2019;1 (2 k) = 0.6; solve for k to get

k = 5.081435. 1 sin Ď&#x20AC;xdx = 2/Ď&#x20AC; 64. (a) 0

1/f 1/f 1 1 2 65. (a) Vrms Vp2 sin2 (2Ď&#x20AC;f t)dt = f Vp2 [1 â&#x2C6;&#x2019; cos(4Ď&#x20AC;f t)]dt = 2 1/f â&#x2C6;&#x2019; 0 0 0 1/f â&#x2C6;&#x161; 1 1 1 2 = f Vp t â&#x2C6;&#x2019; = Vp2 , so Vrms = Vp / 2 sin(4Ď&#x20AC;f t) 4Ď&#x20AC;f 2 2 0

â&#x2C6;&#x161; (b) Vp / 2 = 120, Vp = 120 2 â&#x2030;&#x2C6; 169.7 V

â&#x2C6;&#x161;

us uf

265

66. Let u = t â&#x2C6;&#x2019; x, then du = â&#x2C6;&#x2019;dx and t 0 t f (t â&#x2C6;&#x2019; x)g(x)dx = â&#x2C6;&#x2019; f (u)g(t â&#x2C6;&#x2019; u)du = f (u)g(t â&#x2C6;&#x2019; u)du; 0

the result follows by replacing u by x in the last integral.

a f (a â&#x2C6;&#x2019; u) f (a â&#x2C6;&#x2019; u) + f (u) â&#x2C6;&#x2019; f (u) du = du f (a â&#x2C6;&#x2019; u) + f (u) a f (a â&#x2C6;&#x2019; u) + f (u) 0 a a f (u) du, I = a â&#x2C6;&#x2019; I so 2I = a, I = a/2 du â&#x2C6;&#x2019; = f (a â&#x2C6;&#x2019; u) + f (u) 0 0

67. (a) I = â&#x2C6;&#x2019;

(b) 3/2

1 1 1 1 1 1 2 , dx = â&#x2C6;&#x2019; 2 du, I = (â&#x2C6;&#x2019;1/u )du = â&#x2C6;&#x2019; du = â&#x2C6;&#x2019;I so I = 0 which is 2 2+1 u u 1 + 1/u u â&#x2C6;&#x2019;1 â&#x2C6;&#x2019;1 1 is positive on [â&#x2C6;&#x2019;1, 1]. The substitution u = 1/x is not valid because u impossible because 1 + x2 is not continuous for all x in [â&#x2C6;&#x2019;1, 1].

68. x =

69. (a) Let u = â&#x2C6;&#x2019;x then a f (x)dx = â&#x2C6;&#x2019; â&#x2C6;&#x2019;a

â&#x2C6;&#x2019;a

f (â&#x2C6;&#x2019;u)du =

f (â&#x2C6;&#x2019;u)du = â&#x2C6;&#x2019;

â&#x2C6;&#x2019;a

f (x)dx = 0

â&#x2C6;&#x2019;a

uh a

The graph of f is symmetric about the origin so 0 a a f (x)dx = f (x) dx + f (x)dx = 0 thus

(b)

â&#x2C6;&#x2019;a 0

f (x)dx =

â&#x2C6;&#x2019;a

f (x)dx = â&#x2C6;&#x2019;

f (x)dx is the negative of â&#x2C6;&#x2019;a

f (x)dx, let u = â&#x2C6;&#x2019;x in 0

f (â&#x2C6;&#x2019;u)du = 0

f (x)dx to get â&#x2C6;&#x2019;a

f (u)du = 0

f (x)dx 0

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f (x)dx 0

f (â&#x2C6;&#x2019;u)du = a

f (x)dx + â&#x2C6;&#x2019;a

f (u)du â&#x2C6;&#x2019;a

â&#x2C6;&#x2019;a

so, replacing u by x in the latter integral, a a a f (x)dx = â&#x2C6;&#x2019; f (x)dx, 2 f (x)dx = 0,

266

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Chapter 6

f (x)dx +

f (x)dx = â&#x2C6;&#x2019;a

f (x)dx = 2 0

f (x)dx 0

us uf

The graph of f (x) is symmetric about the y-axis so there is as much signed area to the left of the y-axis as there is to the right. x cos(x2 ) dx = 0

70. (a) By Exercise 69(a), â&#x2C6;&#x2019;1

â&#x2C6;&#x2019;Ď&#x20AC;/2

(b) u = x â&#x2C6;&#x2019; Ď&#x20AC;/2, du = dx, sin(u + Ď&#x20AC;/2) = sin u, cos(u + Ď&#x20AC;/2) = â&#x2C6;&#x2019; sin u Ď&#x20AC; Ď&#x20AC;/2 Ď&#x20AC;/2 sin8 x cos5 x dx = sin8 u(â&#x2C6;&#x2019; sin5 u) du = â&#x2C6;&#x2019; sin13 u du = 0 by Exercise 69(a). â&#x2C6;&#x2019;Ď&#x20AC;/2

EXERCISE SET 6.9 y

(b)

(c)

1 t 1

0.5 1

t 1

3 2

1. (a)

2 1

a/c 1 â&#x2C6;&#x161;a 1

2/a 1

(b) ln t

1/c

= ln(a/c) = 2 â&#x2C6;&#x2019; 5 = â&#x2C6;&#x2019;3

(d) ln t

1 ln a = 9/2 2

(b) ln t

= ln a1/2 = = ln 2 â&#x2C6;&#x2019; 9

(d) ln t

= ln(ac) = ln a + ln c = 7

3 2

3. (a) ln t

t 2 3

a3 1

2a 1

a 2

= ln(1/c) = â&#x2C6;&#x2019;5

= ln a3 = 3 ln a = 6 = ln 2 + 9

= 9 â&#x2C6;&#x2019; ln 2

5. ln 5 â&#x2030;&#x2C6; 1.603210678; ln 5 = 1.609437912; magnitude of error is < 0.0063

uh a

6. ln 3 â&#x2030;&#x2C6; 1.098242635; ln 3 = 1.098612289; magnitude of error is < 0.0004

7. (a) xâ&#x2C6;&#x2019;1 , x > 0 (c) â&#x2C6;&#x2019;x2 , â&#x2C6;&#x2019;â&#x2C6;&#x17E; < x < +â&#x2C6;&#x17E;

(e) x , x > 0 â&#x2C6;&#x161; (g) x â&#x2C6;&#x2019; 3 x, â&#x2C6;&#x2019;â&#x2C6;&#x17E; < x < +â&#x2C6;&#x17E;

(b) x2 , x = 0 (d) â&#x2C6;&#x2019;x, â&#x2C6;&#x2019;â&#x2C6;&#x17E; < x < +â&#x2C6;&#x17E; (f ) ln x + x, x > 0 (h)

ex ,x>0 x

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Exercise Set 6.9

267

8. (a) f (ln 3) = eâ&#x2C6;&#x2019;2 ln 3 = eln(1/9) = 1/9

(b) 2

10. (a) Ď&#x20AC; â&#x2C6;&#x2019;x = eâ&#x2C6;&#x2019;x ln Ď&#x20AC;

â&#x2C6;&#x161; 2

â&#x2C6;&#x161; 2 ln 2

(b) x2x = e2x ln x

yâ&#x2020;&#x2019;0

y 1/3 y 1/3 1 1 1+ 1+ = e1/3 = lim yâ&#x2020;&#x2019;+â&#x2C6;&#x17E; yâ&#x2020;&#x2019;+â&#x2C6;&#x17E; y y 1/3 1/x = lim (1 + x) = e1/3

yâ&#x2020;&#x2019;+â&#x2C6;&#x17E; 1/3x

(b) lim (1 + x) xâ&#x2020;&#x2019;0

1 y

y/3

= lim

12. (a) y = 3x, lim

xâ&#x2020;&#x2019;0

14. g (x) = 1 â&#x2C6;&#x2019; cos x

13. g (x) = x2 â&#x2C6;&#x2019; x

1 =1 x 1 1 sin (b) â&#x2C6;&#x2019; x2 x

3 1 (3x2 ) = x3 x

(b) eln x

â&#x2C6;&#x161; 16. (a) 2x x2 + 1

cos x â&#x2C6;&#x2019;(x2 + 3) sin x â&#x2C6;&#x2019; 2x cos x (x) = , F x2 + 3 (x2 + 3)2

17. F (x) =

(b) 1/3

â&#x2C6;&#x161;

(a) 0 18. F (x) =

x 2 x 2 1 1 1+ 1+ = e2 = lim 11. (a) lim xâ&#x2020;&#x2019;+â&#x2C6;&#x17E; xâ&#x2020;&#x2019;+â&#x2C6;&#x17E; x x 2 1/y 2/y = lim (1 + y) = e2 (b) y = 2x, lim (1 + y)

15. (a)

us uf

9. (a) 3Ď&#x20AC; = eĎ&#x20AC; ln 3

(b) f (ln 2) = eln 2 + 3eâ&#x2C6;&#x2019; ln 2 = 2 + 3eln(1/2) = 2 + 3/2 = 7/2

3x2 + 1, F (x) = â&#x2C6;&#x161;

(a) 0

3x 3x2 + 1 (b)

â&#x2C6;&#x161;

â&#x2C6;&#x161; (c) 6/ 13

d f (t)dt = â&#x2C6;&#x2019; dx

d dx

x2 â&#x2C6;&#x161; d 19. (a) t 1 + tdt = x2 1 + x2 (2x) = 2x3 1 + x2 dx 1 â&#x2C6;&#x161; x2 â&#x2C6;&#x161; 2 2 2 2 4 2 3/2 5/2 (b) t 1 + tdt = â&#x2C6;&#x2019; (x + 1) + (x + 1) â&#x2C6;&#x2019; 3 5 15 1

d dx

uh a

20. (a)

(b)

f (t)dt = â&#x2C6;&#x2019;

g(x)

21. (a) â&#x2C6;&#x2019; sin x2 22. (a) â&#x2C6;&#x2019;(x2 + 1)40

d dx

f (t)dt = â&#x2C6;&#x2019;f (x) a

g(x)

f (t)dt = â&#x2C6;&#x2019;f (g(x))g (x)

tan2 x sec2 x = â&#x2C6;&#x2019; tan2 x 1 + tan2 x 1 1 cos3 (1/x) â&#x2C6;&#x2019; 2 = (b) â&#x2C6;&#x2019; cos3 x x x2 (b) â&#x2C6;&#x2019;

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Chapter 6

3x â&#x2C6;&#x2019; 1 x2 â&#x2C6;&#x2019; 1 + 2x 4 2 9x + 1 x +1

24. If f is continuous on an open interval I and g(x), h(x), and a are in I then g(x) a g(x) h(x) g(x) f (t)dt = f (t)dt + f (t)dt = â&#x2C6;&#x2019; f (t)dt + f (t)dt h(x)

d dx

h(x)

g(x)

f (t)dt = â&#x2C6;&#x2019;f (h(x))h (x) + f (g(x))g (x)

h(x)

1 2 1 (1) â&#x2C6;&#x2019; (â&#x2C6;&#x2019;1) = 1+x 1â&#x2C6;&#x2019;x 1 â&#x2C6;&#x2019; x2

1 1 (3) â&#x2C6;&#x2019; (1) = 0 so F (x) is constant on (0, +â&#x2C6;&#x17E;). F (1) = ln 3 so F (x) = ln 3 for all x > 0. 3x x

f (t)dt = 6,

f (t)dt = 0,

27. from geometry,

f (t)dt = 0; and 5

(4t â&#x2C6;&#x2019; 37)/3dt = â&#x2C6;&#x2019;3

26. F (x) =

f (t)dt

(b)

25. (a) sin2 (x3 )(3x2 ) â&#x2C6;&#x2019; sin2 (x2 )(2x) = 3x2 sin2 (x3 ) â&#x2C6;&#x2019; 2x sin2 (x2 )

23. â&#x2C6;&#x2019;3

us uf

268

(a) F (0) = 0, F (3) = 0, F (5) = 6, F (7) = 6, F (10) = 3 (b) F is increasing where F = f is positive, so on [3/2, 6] and [37/4, 10], decreasing on [0, 3/2] and [6, 37/4]

(d)

(c) critical points when F (x) = f (x) = 0, so x = 3/2, 6, 37/4; maximum 15/2 at x = 6, minimum â&#x2C6;&#x2019;9/4 at x = 3/2

F(x) 6 4 2

-2

f (t)dt =

1 28. fave = 10 â&#x2C6;&#x2019; 0

1 F (10) = 0.3 10

x 1 2 1 (â&#x2C6;&#x2019;t)dt = â&#x2C6;&#x2019; t = (1 â&#x2C6;&#x2019; x2 ), 29. x < 0 : F (x) = 2 2 â&#x2C6;&#x2019;1 â&#x2C6;&#x2019;1 0 x 1 1 x â&#x2030;Ľ 0 : F (x) = (â&#x2C6;&#x2019;t)dt + t dt = + x2 ; F (x) = 2 2 â&#x2C6;&#x2019;1 0 x

uh a

(1 â&#x2C6;&#x2019; x2 )/2, 2

(1 + x )/2,

xâ&#x2030;Ľ0

x 1 30. 0 â&#x2030;¤ x â&#x2030;¤ 2 : F (x) = t dt = x2 , 2 2 0 x x > 2 : F (x) = t dt + 2 dt = 2 + 2(x â&#x2C6;&#x2019; 2) = 2x â&#x2C6;&#x2019; 2; F (x) =

x<0

x2 /2, 0 â&#x2030;¤ x â&#x2030;¤ 2 2x â&#x2C6;&#x2019; 2, x > 2

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Exercise Set 6.9

32. y(x) =

3 t1/3 dt = 2 + t4/3 4

(t1/2 + tâ&#x2C6;&#x2019;1/2 )dt =

x = 1

5 3 4/3 + x 4 4

2 3/2 2 2 8 x â&#x2C6;&#x2019; + 2x1/2 â&#x2C6;&#x2019; 2 = x3/2 + 2x1/2 â&#x2C6;&#x2019; 3 3 3 3

(sec2 t â&#x2C6;&#x2019; sin t)dt = tan x + cos x â&#x2C6;&#x2019;

33. y(x) = 1 +

â&#x2C6;&#x161;

2/2

Ď&#x20AC;/4 x

34. y(x) = 0

2 1 1 te dt = eâ&#x2C6;&#x2019;x â&#x2C6;&#x2019; 2 2

36. s(T ) = s1 +

35. P (x) = P0 +

r(t)dt individuals 0

v(t)dt

us uf

31. y(x) = 2 +

269

37. II has a minimum at x = 12, and I has a zero there, so I could be the derivative of II; on the other hand I has a minimum near x = 1/3, but II is not! zero there, so II could not be the derivative of x I, so I is the graph of f (x) and II is the graph of 0 f (t) dt. 1 k d (x â&#x2C6;&#x2019; 1) = xt t=0 = ln x kâ&#x2020;&#x2019;0 k dt

38. (b) lim

39. (a) where f (t) = 0; by the First Derivative Test, at t = 3 (b) where f (t) = 0; by the First Derivative Test, at t = 1, 5 (c) at t = 0, 1 or 5; from the graph it is evident that it is at t = 5

(d) at t = 0, 3 or 5; from the graph it is evident that it is at t = 3

(e) F is concave up when F = f is positive, i.e. where f is increasing, so on (0, 1/2) and (2, 4); it is concave down on (1/2, 2) and (4, 5) F(x)

(f ) 1 0.5

-0.5

40. (a)

-1

uh a

-4 -2

erf(x)

x 2

-1

(c) erf (x) > 0 for all x, so there are no relative extrema 2 â&#x2C6;&#x161; (e) erf (x) = â&#x2C6;&#x2019;4xeâ&#x2C6;&#x2019;x / Ď&#x20AC; changes sign only at x = 0 so that is the only point of inďŹ&#x201A;ection

(g)

lim erf(x) = +1, lim erf(x) = â&#x2C6;&#x2019;1

xâ&#x2020;&#x2019;+â&#x2C6;&#x17E;

xâ&#x2020;&#x2019;â&#x2C6;&#x2019;â&#x2C6;&#x17E;

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270

Chapter 6

us uf

(a) cos t goes from negative to positive at 2kĎ&#x20AC; â&#x2C6;&#x2019; Ď&#x20AC;/2, and from positive to negative â&#x2C6;&#x161; at t = 2kĎ&#x20AC; + Ď&#x20AC;/2, so C(x) has relative minima when Ď&#x20AC;x2 /2 = 2kĎ&#x20AC; â&#x2C6;&#x2019; Ď&#x20AC;/2, x = Âą â&#x2C6;&#x161;4k â&#x2C6;&#x2019; 1, k = 1, 2, . . ., and C(x) has relative maxima when Ď&#x20AC;x2 /2 = (4k + 1)Ď&#x20AC;/2, x = Âą 4k + 1, k = 0, 1, . . ..

41. C (x) = cos(Ď&#x20AC;x2 /2), C (x) = â&#x2C6;&#x2019;Ď&#x20AC;x sin(Ď&#x20AC;x2 /2)

â&#x2C6;&#x161; (b) sin t changes sign at t = kĎ&#x20AC;, so C(x) has inďŹ&#x201A;ection points at Ď&#x20AC;x2 /2 = kĎ&#x20AC;, x = Âą 2k, k = 1, 2, . . .; the case k = 0 is distinct due to the factor of x in C (x), but x changes sign at x = 0 and sin(Ď&#x20AC;x2 /2) does not, so there is also a point of inďŹ&#x201A;ection at x = 0 x

42. Let F (x) =

x+h

ln tdt; but F (x) = ln x so

x+h

lim

ln tdt = ln x x

provided e3a = 2, a = (ln 2)/3.

43. DiďŹ&#x20AC;erentiate: f (x) = 3e3x , so 2 + a

3e3t dt = 2 + e3t

f (t)dt = 2 + a

= 2 + e3x â&#x2C6;&#x2019; e3a = e3x

1 hâ&#x2020;&#x2019;0 h

F (x + h) â&#x2C6;&#x2019; F (x) 1 = lim ln tdt, F (x) = lim hâ&#x2020;&#x2019;0 hâ&#x2020;&#x2019;0 h h

44. (a) The area under 1/t for x â&#x2030;¤ t â&#x2030;¤ x + 1 is less than the area of the rectangle with altitude 1/x and base 1, but greater than the area of the rectangle with altitude 1/(x + 1) and base 1. x+1 x+1 1 dt = ln t (b) = ln(x + 1) â&#x2C6;&#x2019; ln x = ln(1 + 1/x), so t x x 1/(x + 1) < ln(1 + 1/x) < 1/x for x > 0. (c) from Part (b), e1/(x+1) < eln(1+1/x) < e1/x , e1/(x+1) < 1 + 1/x < e1/x , ex/(x+1) < (1 + 1/x)x < e; by the Squeezing Theorem, lim (1 + 1/x)x = e.

xâ&#x2020;&#x2019;+â&#x2C6;&#x17E;

(d) Use the inequality ex/(x+1) < (1 + 1/x)x to get e < (1 + 1/x)x+1 so (1 + 1/x)x < e < (1 + 1/x)x+1 .

50 1 45. From Exercise 44(d) e â&#x2C6;&#x2019; 1 + < y(50), and from the graph y(50) < 0.06 50

100

0.2

uh a

46. F (x) = f (x), thus F (x) has a value at each x in I because f is continuous on I so F is continuous on I because a function that is diďŹ&#x20AC;erentiable at a point is also continuous at that point

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Chapter 6 Supplementary Exercises

271

CHAPTER 6 SUPPLEMENTARY EXERCISES

5. If the acceleration a = const, then v(t) = at + v0 , s(t) = 12 at2 + v0 t + s0 .

3 1 1 + = 4 2 4 35 3 =â&#x2C6;&#x2019; (c) 5 â&#x2C6;&#x2019;1 â&#x2C6;&#x2019; 4 4

(b) â&#x2C6;&#x2019;1 â&#x2C6;&#x2019;

7. (a)

(d) â&#x2C6;&#x2019;2

(f ) not enough information

(e) not enough information 5 1 +2= 2 2

(b) not enough information

8. (a)

(d) 4(2) â&#x2C6;&#x2019; 3

â&#x2C6;&#x2019;1

(b)

dx +

9. (a)

â&#x2C6;&#x2019;1

1 2 (x + 1)3/2 3

1 â&#x2C6;&#x2019; x2 dx = 2(1) + Ď&#x20AC;(1)2 /2 = 2 + Ď&#x20AC;/2 3 â&#x2C6;&#x2019; Ď&#x20AC;(3)2 /4 = 0

1 (c) u = x , du = 2xdx; 2

1 2

1 (103/2 â&#x2C6;&#x2019; 1) â&#x2C6;&#x2019; 9Ď&#x20AC;/4 3

1 â&#x2C6;&#x2019; u2 du =

1 Ď&#x20AC;(1)2 /4 = Ď&#x20AC;/8 2

10.

13 1 = 2 2

3 1 =â&#x2C6;&#x2019; 2 2

us uf

6. (a) Divide the base into n equal subintervals. Above each subinterval choose the lowest and highest points on the curved top. Draw a rectangle above the subinterval going through the lowest point, and another through the highest point. Add the rectangles that go through the lowest points to obtain a lower estimate of the area; add the rectangles through the highest points to obtain an upper estimate of the area. (b) n = 10: 25.0 cm, 22.4 cm (c) n = 20: 24.4 cm, 23.1 cm

0.8 0.6 0.4 0.2

x 0.2

0.6

11. The rectangle with vertices (0, 0), (Ď&#x20AC;, 0), (Ď&#x20AC;, 1) and (0, 1) has area Ď&#x20AC; and is much too large; so is the triangle with vertices (0, 0), (Ď&#x20AC;, 0) and (Ď&#x20AC;, 1) which has area Ď&#x20AC;/2; 1 â&#x2C6;&#x2019; Ď&#x20AC; is negative; so the answer is 35Ď&#x20AC;/128.

uh a

e2x 3ex 12. Divide ex + 3 into e2x to get x = ex â&#x2C6;&#x2019; x so e +3 e +3 ex e2x dx = ex dx â&#x2C6;&#x2019; 3 dx = ex â&#x2C6;&#x2019; 3 ln(ex + 3) + C x x e +3 e +3

13. Since y = ex and y = ln x are inverse functions, their graphs are symmetric with respect to the line y = x; consequently the areas A1 and A3 are equal (see ďŹ gure). But A1 + A2 = e, so e 1 ln xdx + ex dx = A2 + A3 = A2 + A1 = e 1

y A1

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A3 1

x e

272

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Chapter 6

1 n

(b)

1 n

n k=1

us uf

14. (a)

â&#x2C6;&#x161; k = f (xâ&#x2C6;&#x2014;k )â&#x2C6;&#x2020;x where f (x) = x, xâ&#x2C6;&#x2014;k = k/n, and â&#x2C6;&#x2020;x = 1/n for 0 â&#x2030;¤ x â&#x2030;¤ 1. Thus n k=1 k=1 1 n 1 k 2 lim x1/2 dx = = nâ&#x2020;&#x2019;+â&#x2C6;&#x17E; n n 3 0 n

k=1

k n

4 =

f (xâ&#x2C6;&#x2014;k )â&#x2C6;&#x2020;x where f (x) = x4 , xâ&#x2C6;&#x2014;k = k/n, and â&#x2C6;&#x2020;x = 1/n for 0 â&#x2030;¤ x â&#x2030;¤ 1. Thus

k=1

1 n 4 1 k 1 = x4 dx = lim nâ&#x2020;&#x2019;+â&#x2C6;&#x17E; n n 5 0

k=1

lim

n k=1 n ek/n

nâ&#x2020;&#x2019;+â&#x2C6;&#x17E;

k=1

f (xâ&#x2C6;&#x2014;k )â&#x2C6;&#x2020;x where f (x) = ex , xâ&#x2C6;&#x2014;k = k/n, and â&#x2C6;&#x2020;x = 1/n for 0 â&#x2030;¤ x â&#x2030;¤ 1. Thus = lim

nâ&#x2020;&#x2019;+â&#x2C6;&#x17E;

f (xâ&#x2C6;&#x2014;k )â&#x2C6;&#x2020;x =

k=1

ex dx = e â&#x2C6;&#x2019; 1. 0

(c)

k=1

n ek/n

1 is positive and increasing on the interval [1, 2], the left endpoint approximation x 1 overestimates the integral of and the right endpoint approximation underestimates it. x (a) For n = 5 this becomes 2 1 1 1 1 1 1 1 1 1 1 1 < 0.2 + + + + dx < 0.2 + + + + 1.2 1.4 1.6 1.8 2.0 1.0 1.2 1.4 1.6 1.8 1 x 2 1 dx = ln 2 is (b) For general n the left endpoint approximation to x 1 n n nâ&#x2C6;&#x2019;1 1 1 1 1 = = and the right endpoint approximation is n 1 + (k â&#x2C6;&#x2019; 1)/n n+kâ&#x2C6;&#x2019;1 n+k k=1 k=1 k=0 2 n n nâ&#x2C6;&#x2019;1 1 1 1 1 . This yields < dx < which is the desired inequality. n+k n+k n+k 1 x k=1

k=1

15. Since f (x) =

k=0

1 1 1 1 = so â&#x2030;¤ 0.1, n â&#x2030;Ľ 5 (c) By telescoping, the diďŹ&#x20AC;erence is â&#x2C6;&#x2019; n 2n 2n 2n (d) n â&#x2030;Ľ 1, 000

16. The direction ďŹ eld is clearly an even function, which means that the solution is even, its derivative is odd. Since sin x is periodic and the direction ďŹ eld is not, that eliminates all but x, the solution of which is the family y = x2 /2 + C.

17. (a) 1 Âˇ 2 + 2 Âˇ 3 + Âˇ Âˇ Âˇ + n(n + 1) =

n k=1

k(k + 1) =

n k=1

k2 +

k=1

uh a

1 1 1 n(n + 1)(2n + 1) + n(n + 1) = n(n + 1)(n + 2) 6 2 3 nâ&#x2C6;&#x2019;1 nâ&#x2C6;&#x2019;1 nâ&#x2C6;&#x2019;1 9 9 1 9 k 1 1 17 n â&#x2C6;&#x2019; 1 ; = 1â&#x2C6;&#x2019; 2 k = (n â&#x2C6;&#x2019; 1) â&#x2C6;&#x2019; 2 Âˇ (n â&#x2C6;&#x2019; 1)(n) = â&#x2C6;&#x2019; (b) n n2 n n n n 2 2 n k=1 k=1 k=1 17 17 n â&#x2C6;&#x2019; 1 lim = nâ&#x2020;&#x2019;+â&#x2C6;&#x17E; 2 n 2 ďŁŽ ďŁš 3 2 2 3 3 3 1 1 ďŁ° ďŁť 2i + (2)(3) = 2 (c) i+ j = i+ 3 = 2 Âˇ (3)(4) + (3)(3) = 21 2 2 i=1 j=1 j=1 i=1 i=1 i=1 =

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Chapter 6 Supplementary Exercises

18. (a)

(k + 4)(k + 1)

(b)

(k â&#x2C6;&#x2019; 1)(k â&#x2C6;&#x2019; 4)

k=5

k=0

273

us uf

19. For 1 â&#x2030;¤ k â&#x2030;¤ n the k-th L-shaped strip consists of the corner square, a strip above and a strip to n the right for a combined area of 1 + (k â&#x2C6;&#x2019; 1) + (k â&#x2C6;&#x2019; 1) = 2k â&#x2C6;&#x2019; 1, so the total area is (2k â&#x2C6;&#x2019; 1) = n2 . k=1 n

(2k â&#x2C6;&#x2019; 1) = 2

k=1

kâ&#x2C6;&#x2019;

k=1

n k=1

1 1 = 2 Âˇ n(n + 1) â&#x2C6;&#x2019; n = n2 2

21. (35 â&#x2C6;&#x2019; 34 ) + (36 â&#x2C6;&#x2019; 35 ) + Âˇ Âˇ Âˇ + (317 â&#x2C6;&#x2019; 316 ) = 317 â&#x2C6;&#x2019; 34 1 1 1 50 1 1 + â&#x2C6;&#x2019; + ÂˇÂˇÂˇ + = 22. 1â&#x2C6;&#x2019; â&#x2C6;&#x2019; 2 2 3 50 51 51 1 1 1 1 1 1 1 399 23. + + Âˇ Âˇ Âˇ + = 2 â&#x2C6;&#x2019;1=â&#x2C6;&#x2019; â&#x2C6;&#x2019; â&#x2C6;&#x2019; â&#x2C6;&#x2019; 12 32 22 202 192 20 400 22

20. 1 + 3 + 5 + Âˇ Âˇ Âˇ + (2n â&#x2C6;&#x2019; 1) =

24. (22 â&#x2C6;&#x2019; 2) + (23 â&#x2C6;&#x2019; 22 ) + Âˇ Âˇ Âˇ + (2101 â&#x2C6;&#x2019; 2100 ) = 2101 â&#x2C6;&#x2019; 2

n 1 1 1 1 = â&#x2C6;&#x2019; 25. (a) (2k â&#x2C6;&#x2019; 1)(2k + 1) 2 2k â&#x2C6;&#x2019; 1 2k + 1 k=1 k=1 1 1 1 1 1 1 1 1 = + + + ÂˇÂˇÂˇ + 1â&#x2C6;&#x2019; â&#x2C6;&#x2019; â&#x2C6;&#x2019; â&#x2C6;&#x2019; 3 3 5 5 7 2n â&#x2C6;&#x2019; 1 2n + 1 2 1 1 n = 1â&#x2C6;&#x2019; = 2 2n + 1 2n + 1 1 n = (b) lim nâ&#x2020;&#x2019;+â&#x2C6;&#x17E; 2n + 1 2 n n 1 1 1 = â&#x2C6;&#x2019; 26. (a) k(k + 1) k k+1 k=1 k=1 1 1 1 1 1 1 1 = 1â&#x2C6;&#x2019; + â&#x2C6;&#x2019; + â&#x2C6;&#x2019; + ÂˇÂˇÂˇ + â&#x2C6;&#x2019; 2 2 3 3 4 n n+1

27.

lim

nâ&#x2020;&#x2019;+â&#x2C6;&#x17E;

(xi â&#x2C6;&#x2019; x ÂŻ) =

i=1

xi â&#x2C6;&#x2019;

i=1

n =1 n+1

(b)

xi = nÂŻ x so

uh a

i=1

28. S â&#x2C6;&#x2019; rS =

n 1 = n+1 n+1

=1â&#x2C6;&#x2019;

k=0

x ÂŻ=

i=1

1 xi thus n i=1 n

xi â&#x2C6;&#x2019; nÂŻ x but x ÂŻ=

i=1

(xi â&#x2C6;&#x2019; x ÂŻ) = nÂŻ x â&#x2C6;&#x2019; nÂŻ x=0

i=1

ark â&#x2C6;&#x2019;

ark+1

k=0

= (a + ar + ar2 + Âˇ Âˇ Âˇ + arn ) â&#x2C6;&#x2019; (ar + ar2 + ar3 + Âˇ Âˇ Âˇ + arn+1 ) = a â&#x2C6;&#x2019; arn+1 = a(1 â&#x2C6;&#x2019; rn+1 )

so (1 â&#x2C6;&#x2019; r)S = a(1 â&#x2C6;&#x2019; rn+1 ), hence S = a(1 â&#x2C6;&#x2019; rn+1 )/(1 â&#x2C6;&#x2019; r)

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Chapter 6

3k+1 =

k=0

(b)

(c)

3 3(1 â&#x2C6;&#x2019; 320 ) = (320 â&#x2C6;&#x2019; 1) 1â&#x2C6;&#x2019;3 2

3(3k ) =

k=0

2k+5 =

k=0 100

(â&#x2C6;&#x2019;1)

k=0

25 2k =

k=0

â&#x2C6;&#x2019;1 2

k =

25 (1 â&#x2C6;&#x2019; 226 ) = 231 â&#x2C6;&#x2019; 25 1â&#x2C6;&#x2019;2

us uf

29. (a)

(â&#x2C6;&#x2019;1)(1 â&#x2C6;&#x2019; (â&#x2C6;&#x2019;1/2)101 ) 2 = â&#x2C6;&#x2019; (1 + 1/2101 ) 1 â&#x2C6;&#x2019; (â&#x2C6;&#x2019;1/2) 3

30. (a) 1.999023438, 1.999999046, 2.000000000; 2

(b) 2.831059456, 2.990486364, 2.999998301; 3

274

31. (a) If u = sec x, du = sec x tan xdx, sec x tan xdx = udu = u2 /2 + C1 = (sec2 x)/2 + C1 ; if u = tan x, du = sec2 xdx, sec2 x tan xdx = udu = u2 /2 + C2 = (tan2 x)/2 + C2 .

(b) They are equal only if sec2 x and tan2 x diďŹ&#x20AC;er by a constant, which is true.

n b

fk (x)dx =

a k=1

n k=1

fk (x)dx

34. (a)

Ď&#x20AC;/4 Ď&#x20AC;/4 1 1 1 2 32. sec x = (2 â&#x2C6;&#x2019; 1) = 1/2 and 12 tan2 x = (1 â&#x2C6;&#x2019; 0) = 1/2 2 2 2 0 0 2 33. 1 + xâ&#x2C6;&#x2019;2/3 dx = xâ&#x2C6;&#x2019;1/3 x2/3 + 1dx; u = x2/3 + 1, du = xâ&#x2C6;&#x2019;1/3 dx 3 3 u1/2 du = u3/2 + C = (x2/3 + 1)3/2 + C 2

(b) yes; substitute ck fk (x) for fk (x) in part (a), and then use

ck fk (x)dx = ck a

fk (x)dx a

from Theorem 6.5.4

35. left endpoints: xâ&#x2C6;&#x2014;k = 1, 2, 3, 4;

f (xâ&#x2C6;&#x2014;k )â&#x2C6;&#x2020;x = (2 + 3 + 2 + 1)(1) = 8

k=1

right endpoints: xâ&#x2C6;&#x2014;k = 2, 3, 4, 5;

f (xâ&#x2C6;&#x2014;k )â&#x2C6;&#x2020;x = (3 + 2 + 1 + 2)(1) = 8

k=1

36. (a) xâ&#x2C6;&#x2014;k = 0, 1, 2, 3, 4

k=1

f (xâ&#x2C6;&#x2014;k )â&#x2C6;&#x2020;x = e0 + e1 + e2 + e3 + e4 (1) = (1 â&#x2C6;&#x2019; e5 )/(1 â&#x2C6;&#x2019; e) = 85.791

(b) xâ&#x2C6;&#x2014;k = 1, 2, 3, 4, 5 4

f (xâ&#x2C6;&#x2014;k )â&#x2C6;&#x2020;x = e1 + e2 + e3 + e4 + e5 (1) = e(1 â&#x2C6;&#x2019; e5 )/(1 â&#x2C6;&#x2019; e) = 233.204

uh a

k=1

(c) xâ&#x2C6;&#x2014;k = 1/2, 3/2, 5/2, 7/2, 9/2 4 f (xâ&#x2C6;&#x2014;k )â&#x2C6;&#x2020;x = e1/2 + e3/2 + e5/2 + e7/2 + e9/2 (1) = e1/2 (1 â&#x2C6;&#x2019; e5 )/(1 â&#x2C6;&#x2019; e) = 141.446

k=1

37. fave =

1 eâ&#x2C6;&#x2019;1

e 1 1 1 1 1 dx = ln x = ; â&#x2C6;&#x2014; = , xâ&#x2C6;&#x2014; = e â&#x2C6;&#x2019; 1 x eâ&#x2C6;&#x2019;1 e â&#x2C6;&#x2019; 1 x e â&#x2C6;&#x2019; 1 1

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Chapter 6 Supplementary Exercises

39. 0.351220577, 0.420535296, 0.386502483

40. 1.63379940, 1.805627583, 1.717566087

1 1 41. f (x) = e , [a, b] = [0, 1], â&#x2C6;&#x2020;x = ; lim f (xâ&#x2C6;&#x2014;k ) = n nâ&#x2020;&#x2019;+â&#x2C6;&#x17E; n n

ex dx = e â&#x2C6;&#x2019; 1

k=1

42. (a) ex

44. (a) F (x) = (b) F (x) =

(b) â&#x2C6;&#x2019; ln(e2 +eâ&#x2C6;&#x2019;1)

1 dt 1 + et

â&#x2C6;&#x161; â&#x2C6;&#x161; â&#x2C6;&#x161; â&#x2C6;&#x161; x2 â&#x2C6;&#x2019; 3 ; increasing on (â&#x2C6;&#x2019;â&#x2C6;&#x17E;, â&#x2C6;&#x2019; 3], [ 3, +â&#x2C6;&#x17E;), decreasing on [â&#x2C6;&#x2019; 3, 3] x2 + 7 20x ; concave down on (â&#x2C6;&#x2019;â&#x2C6;&#x17E;, 0), concave up on (0, +â&#x2C6;&#x17E;) (x2 + 7)2

lim F (x) = â&#x2C6;&#x201C;â&#x2C6;&#x17E;, so F has no absolute extrema.

xâ&#x2020;&#x2019;Âąâ&#x2C6;&#x17E;

(d)

0.5

(c)

1 dt 1 + et

43. (a)

(b) ln x

k=1

us uf

lim

nâ&#x2020;&#x2019;+â&#x2C6;&#x17E;

25(k â&#x2C6;&#x2019; 1)2 5 125 â&#x2C6;&#x2019; = n2 n 6

38.

n 25(k â&#x2C6;&#x2019; 1)

275

F(x) x 4

-4

45. F (x) =

-0.5

1 1 + (â&#x2C6;&#x2019;1/x2 ) = 0 so F is constant on (0, +â&#x2C6;&#x17E;). 2 1+x 1 + (1/x)2

46. (â&#x2C6;&#x2019;3, 3) because f is continuous there and 1 is in (â&#x2C6;&#x2019;3, 3)

47. (a) The domain is (â&#x2C6;&#x2019;â&#x2C6;&#x17E;, +â&#x2C6;&#x17E;); F (x) is 0 if x = 1, positive if x > 1, and negative if x < 1, because the integrand is positive, so the sign of the integral depends on the orientation (forwards or backwards). (b) The domain is [â&#x2C6;&#x2019;2, 2]; F (x) is 0 if x = â&#x2C6;&#x2019;1, positive if â&#x2C6;&#x2019;1 < x â&#x2030;¤ 2, and negative if â&#x2C6;&#x2019;2 â&#x2030;¤ x < â&#x2C6;&#x2019;1; same reasons as in Part (a).

uh a

48. The left endpoint of the top boundary is ((b â&#x2C6;&#x2019; a)/2, h) and the right endpoint of the top boundary is ((b + a)/2, h) so ďŁą x < (b â&#x2C6;&#x2019; a)/2 ďŁ˛ 2hx/(b â&#x2C6;&#x2019; a), h, (b â&#x2C6;&#x2019; a)/2 < x < (b + a)/2 f (x) = ďŁł 2h(x â&#x2C6;&#x2019; b)/(a â&#x2C6;&#x2019; b), x > (a + b)/2 The area of the trapezoid is given by (bâ&#x2C6;&#x2019;a)/2 (b+a)/2 b 2hx 2h(x â&#x2C6;&#x2019; b) dx+ dx = (bâ&#x2C6;&#x2019;a)h/4+ah+(bâ&#x2C6;&#x2019;a)h/4 = h(a+b)/2. hdx+ bâ&#x2C6;&#x2019;a aâ&#x2C6;&#x2019;b 0 (bâ&#x2C6;&#x2019;a)/2 (b+a)/2

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Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Chapter 6

no, since the velocity curve is not a straight line 25 < t < 40 (c) 3.54 ft/s (d) 141.5 ft no since the velocity is positive and the acceleration is never negative need the position at any one given time (e.g. s0 )

t 2

50. w( t) =

Ď&#x201E; /7 dĎ&#x201E; = t /14, assuming w0 = w(0) = 0; wave 0

1 = 26

1 t3 t /7 dt = 26 21 2

26 â&#x2C6;&#x161; 21, so t â&#x2030;&#x2C6; 39.716, so during the 40th week. 3 1 1â&#x2C6;&#x161; 1 â&#x2C6;&#x161; du = u1/2 + C = 5 + 2 sin 3x + C u = 5 + 2 sin 3x, du = 6 cos 3xdx; 3 3 6 u â&#x2C6;&#x161; â&#x2C6;&#x161; â&#x2C6;&#x161; 4 4 1 u = 3 + x, du = â&#x2C6;&#x161; dx; 2 udu = u3/2 + C = (3 + x)3/2 + C 3 3 2 x 1 1 1 u = ax3 + b, du = 3ax2 dx; du = â&#x2C6;&#x2019; +C =â&#x2C6;&#x2019; 2 3 +C 3au2 3au 3a x + 3ab 1 1 1 2 sec2 udu = tan u + C = tan(ax2 ) + C u = ax , du = 2axdx; 2a 2a 2a

54.

55.

1 3 1 â&#x2C6;&#x2019; 3â&#x2C6;&#x2019; + 4 3u u 4u

â&#x2C6;&#x2019;1 = 389/192

57. u = ln x, du = (1/x)dx; 1

58.

1 du = ln u u

â&#x2C6;&#x161; eâ&#x2C6;&#x2019;x/2 dx = 2(1 â&#x2C6;&#x2019; 1/ e)

59. u = e

â&#x2C6;&#x2019;2x

, du = â&#x2C6;&#x2019;2e

1 1 3 sin Ď&#x20AC;x = 0 3Ď&#x20AC; 0

= ln 2 1

â&#x2C6;&#x2019;2x

56.

â&#x2C6;&#x2019;2

53.

= 676/3

52.

51.

Set 676/3 = t2 /14, t = Âą

49. (a) (b) (e) (f )

us uf

276

1 dx; â&#x2C6;&#x2019; 2

3 1 (1 + cos u)du = + 8 2

100,000

60. 100,000/(ln 100,000) â&#x2030;&#x2C6; 8686;

1 sin 1 â&#x2C6;&#x2019; sin 4

1 dt â&#x2030;&#x2C6; 9629, so the integral is better ln t

1/4

(x + x2 â&#x2C6;&#x2019; x3 )dx = 1.007514.

61. With b = 1.618034, area =

1 2 2 2 x sin 3x â&#x2C6;&#x2019; sin 3x + x cos 3x â&#x2C6;&#x2019; 0.251607 3 27 9 4 (b) f (x) = 4 + x2 + â&#x2C6;&#x161; â&#x2C6;&#x2019;6 4 + x2

62. (a) f (x) =

7 1 4 k â&#x2C6;&#x2019; k â&#x2C6;&#x2019; k 2 + = 0 to get k = 2.073948. 4 4 1 3 1 1 (b) Solve â&#x2C6;&#x2019; cos 2k + k + = 3 to get k = 1.837992. 2 3 2 x t x â&#x2C6;&#x161; 64. F (x) = dt, F (x) = â&#x2C6;&#x161; , so F is increasing on [1, 3]; Fmax = F (3) â&#x2030;&#x2C6; 1.152082854 3 2+t 2 + x3 â&#x2C6;&#x2019;1 and Fmin = F (1) â&#x2030;&#x2C6; â&#x2C6;&#x2019;0.07649493141

uh a

63. (a) Solve

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Chapter 6 Horizon Module

(b)

0.7651976866

(c)

J0 (x) = 0 if x = 2.404826

y = J0(x)

0.5

x 1 2

6 7 8

65. (a)

277

us uf

-0.5

CHAPTER 6 HORIZON MODULE

1. vx (0) = 35 cos α, so from Equation (1), x(t) = (35 cos α)t; vy (0) = 35 sin α, so from Equation (2), y(t) = (35 sin α)t − 4.9t2 . dy(t) dx(t) = 35 sin α − 9.8t = 35 cos α, vy (t) = dt dt (b) vy (t) = 35 sin α − 9.8t, vy (t) = 0 when t = 35 sin α/9.8; y = vy (0)t − 4.9t2 = (35 sin α)(35 sin α)/9.8 − 4.9((35 sin α)/9.8)2 = 62.5 sin2 α, so ymax = 62.5 sin2 α.

2. (a) vx (t) =

15◦ no

25◦ yes

35◦ no

45◦ no

55◦ no

65◦ yes

75◦ no

85◦ no

120 0

0.004 2 x ; 3. t = x/(35 cos α) so y = (35 sin α)(x/(35 cos α)) − 4.9(x/(35 cos α))2 = (tan α)x − cos2 α the trajectory is a parabola because y is a quadratic function of x.

5. y(t) = (35 sin α s)t − 4.9t2 = 0 when t = 35 sin α/4.9, at which time x = (35 cos α)(35 sin α/4.9) = 125 sin 2α; this is the maximum value of x, so R = 125 sin 2α m. 6. (a) R = 95 when sin 2α = 95/125 = 0.76, α = 0.4316565575, 1.139139769 rad ≈ 24.73◦ , 65.27◦ . (b) y(t) < 50 is required; but y(1.139) ≈ 51.56 m, so his height would be 56.56 m.

uh a

7. 0.4019 < α < 0.4636 (radians), or 23.03◦ < α < 26.57◦

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Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH CHAPTER 7

us uf

Applications of the Definite Integral in Geometry, Science, and Engineering EXERCISE SET 7.1

−1 4

2. A =

−1

4 √ 3/2 2 ( x + x/4)dx = (2x /3 + x /8) = 22/3

0 2

2 (y − 1/y )dy = (y /2 + 1/y) = 1 2

3. A =

4. A =

1 2

2 (x2 + 1 − x)dx = (x3 /3 + x − x2 /2) = 9/2

2 (2 − y 2 + y)dy = (2y − y 3 /3 + y 2 /2) = 10/3

4 2

(4x − x )dx = 32/3

5. (a) A =

1. A =

(b) A =

√ ( y − y/4)dy = 32/3

(4, 16) y = 4x y = x2 5

(a) A =

6. Eliminate x to get y 2 = 4(y + 4)/2, y 2 − 2y − 8 = 0, (y − 4)(y + 2) = 0; y = −2, 4 with corresponding values of x = 1, 4. √ √ [2 x − (−2 x)]dx +

√

4 xdx +

√ [2 x − (2x − 4)]dx

y = 2x – 4 x

√ (2 x − 2x + 4)dx = 8/3 + 19/3 = 9

(1, -2)

(b) A =

[(y/2 + 2) − y 2 /4]dy = 9

uh a

−2

(4, 4)

y2 = 4x

x 1

278

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Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Exercise Set 7.1

√ ( x − x2 )dx = 49/192

7. A =

279

y = x2

us uf

(1, 1)

y = √x

1/4

2 3

[0 − (x − 4x)]dx

8. A =

(0 − cos 2x)dx

0 2

π/4

(4x − x3 )dx = 4

π/2

9. A =

cos 2x dx = 1/2 π/4

y 1

y = cos 2 x

2 -1

y = 2x3 – 4x

10. Equate sec2 x and 2 to get sec2 x = 2,

y = sec2 x x

(3, 2)

12. A =

x = sin y 9

[(x + 2) − x2 ]dx = 9/2

y (2, 4)

(–1, 1)

y = x2 x

x=y–2

uh a

−1

√

π/4

−π/4

sin y dy =

√ sec x = ± 2, x = ±π/4 π/4 A= (2 − sec2 x)dx = π − 2

3π/4

11. A =

y 2

(#, 2)

π/2

=−

1 4

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Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Chapter 7

e2x − ex dx 0 ln 2 1 2x = e − ex = 1/2 2

14. A = 1

e dy = ln y = 1 y 1

ln 2

13. A =

us uf

280

y y = e2x

x 1/e y = ex

ln 2

16.

√ 3 1 √ = 2, x = ± , so 2 2 1−x √ 3/2 1 dx 2− √ A= √ 1 − x2 − 3/2 √3/2 √ −1 = 2 3 − 23 π = 2 − sin x √

15.

2 A= − |x| dx 1 + x2 −1 1 2 =2 − x dx 1 + x2 0 1 = 4 tan−1 x − x2 = π − 1

− 3/2

-1

3 − x, x ≤ 1 , 1 + x, x ≥ 1 1 1 A= − x + 7 − (3 − x) dx 5 −5 5 1 + − x + 7 − (1 + x) dx 5 1 1 5 4 6 6 − x dx = x + 4 dx + 5 5 −5 1

1.5 1 y=

1 1 - x2

0.5 x

3 2

y (–5, 8) y = – 15 x + 7 (5, 6) y = 3–x

y = 1+x

17. y = 2 + |x − 1| =

y=2

uh a

= 72/5 + 48/5 = 24

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Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Exercise Set 7.1

2/5

(4x − x)dx

18. A =

19. A =

(−x + 2 − x)dx

+ 2/5

(x3 − 4x2 + 3x)dx 3 + [−(x3 − 4x2 + 3x)]dx 0

2/5

(2 − 2x)dx = 3/5

3x dx + 0

= 5/12 + 32/12 = 37/12

2/5

( 25 , 85 ) y = 4x

-1

y = -x + 2 (1, 1) x

-8

(x3 − 4x2 + 3x)dx +

= =

8 37 5 + = 12 3 12

Ri -2

22. The region is symmetric about the origin so 2 |x3 − 4x|dx = 8 A=2 0

3.1

-3

-3.1

uh a

-1

21. From the symmetry of the region 5π/4 √ A=2 (sin x − cos x)dx = 4 2 π/4

(−x3 + 4x2 − 3x)dx 1

-1

[(2x3 − 3x) − (x3 − 2x2 )]dx

y= x

20. Equate y = x3 − 2x2 and y = 2x2 − 3x to get x3 − 4x2 + 3x = 0, x(x − 1)(x − 3) = 0; x = 0, 1, 3 with corresponding values of y = 0, −1.9. 1 [(x3 − 2x2 ) − (2x2 − 3x)]dx A=

us uf

281

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Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Chapter 7

â&#x2C6;&#x2019;1

(y 3 â&#x2C6;&#x2019; y)dy +

â&#x2C6;&#x2019;(y 3 â&#x2C6;&#x2019; y)dy

y â&#x2C6;&#x2019; 4y 2 + 3y â&#x2C6;&#x2019; (y 2 â&#x2C6;&#x2019; y) dy

24. A =

= 1/2

y â&#x2C6;&#x2019; y â&#x2C6;&#x2019; (y 3 â&#x2C6;&#x2019; 4y 2 + 3y) dy

= 7/12 + 45/4 = 71/6 4.1 1

-1

-1 -2.2

12.1

â&#x2C6;&#x161;

k = 0.997301.

â&#x2C6;&#x161; â&#x2C6;&#x161; 2 2 â&#x2C6;&#x2019;1 = 4 2 â&#x2C6;&#x2019; 2 sin 3

â&#x2C6;&#x161; eâ&#x2C6;&#x2019;2 2/3

1.5 1 0.5 x 0.5

y 20 15 10 5 x 1

(1/ 1 â&#x2C6;&#x2019; x2 â&#x2C6;&#x2019; x)dx = sinâ&#x2C6;&#x2019;1 k â&#x2C6;&#x2019; k 2 /2 = 1; solve for k to get

27. The area is given by

2/3

= 3 ln x â&#x2C6;&#x2019; sinâ&#x2C6;&#x2019;1 (ln x)

â&#x2C6;&#x161;

26. The curves meet for x = eâ&#x2C6;&#x2019;2 2/3 , e2 2/3 thus e2â&#x2C6;&#x161;2/3 1 3 â&#x2C6;&#x2019; A= dx â&#x2C6;&#x161; x x 1 â&#x2C6;&#x2019; (ln x)2 eâ&#x2C6;&#x2019;2 2/3 e2

2.5

â&#x2C6;&#x161; 25. The curves meet when x = ln 2, so â&#x2C6;&#x161;ln 2 â&#x2C6;&#x161;ln 2 2 2 1 1 (2x â&#x2C6;&#x2019; xex ) dx = x2 â&#x2C6;&#x2019; ex = ln 2 â&#x2C6;&#x2019; A= 2 2 0 0

23. A =

us uf

282

28. The b curves intersect at x = a = 0 and x = b = 0.838422 so the area is (sin 2x â&#x2C6;&#x2019; sinâ&#x2C6;&#x2019;1 x)dx â&#x2030;&#x2C6; 0.174192. 29. Solve 3â&#x2C6;&#x2019;2x = x6 +2x5 â&#x2C6;&#x2019;3x4 +x2 to ďŹ nd the real roots x = â&#x2C6;&#x2019;3, 1; from a plot it is seen that the line 1

â&#x2C6;&#x2019;3

(3â&#x2C6;&#x2019;2xâ&#x2C6;&#x2019;(x6 +2x5 â&#x2C6;&#x2019;3x4 +x2 )) dx = 9152/105

uh a

is above the polynomial when â&#x2C6;&#x2019;3 < x < 1, so A =

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√ 2 ydy =

√ 2 ydy

y 1/2 dy k

k3 = 4 √ k= 34

2 2 3/2 = (27 − k 3/2 ) k 3 3 k 3/2 = 27/2 2/3

k = (27/2)

x2 dx

1 3 1 k = (8 − k 3 ) 3 3

y 1/2 dy =

x2 dx =

32.

31.

√ 3

x = √y

= 9/ 4

us uf

√ 1 30. Solve x5 − 2x3 − 3x = x3 to ﬁnd the roots x = 0, ± 6 + 2 21. Thus, by symmetry, 2 √(6+2√21)/2 7√ 27 + (x3 − (x5 − 2x3 − 3x)) dx = 21 A=2 4 4 0

283

Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Exercise Set 7.1

y=9 y=k

x=k

(2x − x2 )dx = 4/3

33. (a) A = 0

(b) y = mx intersects y = 2x − x2 where mx = 2x − x2 , x2 + (m − 2)x = 0, x(x + m − 2) = 0 so x = 0 or x = 2 − m. The area below the curve and above the line is 2−m

2−m 2−m 1 1 1 (2 − m)x2 − x3 (2x − x2 − mx)dx = [(2 − m)x − x2 ]dx = = (2 − m)3 2 3 6 0 0 0 √ 3 3 3 so (2 − m) /6 = (1/2)(4/3) = 2/3, (2 − m) = 4, m = 2 − 4.

34. The line through (0, 0) and (5π/6, 1/2) is y = A= 0

√ 5 3 3 x dx = − π+1 sin x − 5π 2 24

y 1

y = sin x

5π/6

3 x; 5π

( 56c , 12 ) c

35. (a) It gives the area of the region that is between f and g when f (x) > g(x) minus the area of the region between f and g when f (x) < g(x), for a ≤ x ≤ b.

uh a

(b) It gives the area of the region that is between f and g for a ≤ x ≤ b.

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Chapter 7

36. (b)

1 1/n

lim

nâ&#x2020;&#x2019;+â&#x2C6;&#x17E;

â&#x2C6;&#x2019; x) dx = lim

nâ&#x2020;&#x2019;+â&#x2C6;&#x17E;

n x2 x(n+1)/n â&#x2C6;&#x2019; n+1 2

= lim

nâ&#x2020;&#x2019;+â&#x2C6;&#x17E;

1 n â&#x2C6;&#x2019; n+1 2

= 1/2

us uf

37. The curves intersect at x = 0 and, by Newtonâ&#x20AC;&#x2122;s Method, at x â&#x2030;&#x2C6; 2.595739080 = b, so b

b (sin x â&#x2C6;&#x2019; 0.2x)dx = â&#x2C6;&#x2019; cos x + 0.1x2 â&#x2030;&#x2C6; 1.180898334 Aâ&#x2030;&#x2C6;

284

38. By Newtonâ&#x20AC;&#x2122;s Method, the points of intersection are at x â&#x2030;&#x2C6; Âą0.824132312, so with b b 2 3 b = 0.824132312 we have A â&#x2030;&#x2C6; 2 (cos x â&#x2C6;&#x2019; x )dx = 2(sin x â&#x2C6;&#x2019; x /3) â&#x2030;&#x2C6; 1.094753609 0

39. By Newtonâ&#x20AC;&#x2122;s Method the points of intersection are x = x1 â&#x2030;&#x2C6; 0.4814008713 and x2 ln x â&#x2C6;&#x2019; (x â&#x2C6;&#x2019; 2) dx â&#x2030;&#x2C6; 1.189708441. x = x2 â&#x2030;&#x2C6; 2.363938870, and A â&#x2030;&#x2C6; x x1

|v| dt, so 60 (3t â&#x2C6;&#x2019; t2 /20) dt = 1800 ft. (a) distance = 0

(3t â&#x2C6;&#x2019; t2 /20) dt =

(b) If T â&#x2030;¤ 60 then distance =

42. Since a1 (0) = a2 (0) = 0, A =

(a2 (t)â&#x2C6;&#x2019;a1 (t)) dt = v2 (T )â&#x2C6;&#x2019;v1 (T ) is the diďŹ&#x20AC;erence in the velocities 0

of the two cars at time T .

3 2 1 T â&#x2C6;&#x2019; T 3 ft. 2 60

41. distance =

40. By Newtonâ&#x20AC;&#x2122;s of intersection are x = Âąx1 where x1 â&#x2030;&#x2C6; 0.6492556537, thus x1 Method the points 2 â&#x2C6;&#x2019; 3 + 2 cos x dx â&#x2030;&#x2C6; 0.826247888 Aâ&#x2030;&#x2C6;2 1 + x2 0

43. Solve x1/2 + y 1/2 = a1/2 for y to get

y = (a1/2 â&#x2C6;&#x2019; x1/2 )2 = a â&#x2C6;&#x2019; 2a1/2 x1/2 + x a A= (a â&#x2C6;&#x2019; 2a1/2 x1/2 + x)dx = a2 /6 0

of the ellipse; make use of symmetry to 1 2 Ď&#x20AC;a = Ď&#x20AC;ab. 4

â&#x2C6;&#x161; 44. Solve for y to get y = (b/a) a2 â&#x2C6;&#x2019; x2 for the upper half a b 4b a 2 4b 2 2 a â&#x2C6;&#x2019; x dx = a â&#x2C6;&#x2019; x2 dx = get A = 4 Âˇ a 0 a 0 a

x a

uh a

45. Let A be the area between the curve and the x-axis and AR the area of the rectangle, then b b k kbm+1 A= xm+1 = , AR = b(kbm ) = kbm+1 , so A/AR = 1/(m + 1). kxm dx = m + 1 m + 1 0 0

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Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Exercise Set 7.2

285

EXERCISE SET 7.2

1. V = π −1

(3 − x)dx = 8π

[(2 − x2 )2 − x2 ]dx

2. V = π 0

(4 − 5x2 + x4 )dx

=π 0

= 38π/15 2

1 (3 − y)2 dy = 13π/6 4

4. V = π

x4 dx = 32π/5

6. V = π

(4 − 1/y 2 )dy = 9π/2 1/2

5. V = π

3. V = π

π/3

√ sec2 x dx = π( 3 − 1)

π/4

us uf

y = sec x

y = x2

2 1

-1

π/2

cos x dx = (1 −

7. V = π

√

2/2)π

π/4

y = √cos x x 6

-1

−4

(x4 − x6 )dx = 2π/35

y 1

(1, 1) y = x2 y = x3

10. V = π −3

(9 − x2 )2 dx

=π −3

(81 − 18x2 + x4 )dx = 1296π/5 y

y = √25 – x2

9 y = 9 – x2

y=3 x

x -3

uh a

[(x2 )2 − (x3 )2 ]dx

(16 − x2 )dx = 256π/3

=π

= 2π

[(25 − x ) − 9]dx

9. V = π

8. V = π

-2

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Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Chapter 7

4 2

2 2

[(4x) − (x ) ]dx

11. V = π 0

π/4

(cos2 x − sin2 x)dx

12. V = π 0

(16x2 − x4 )dx = 2048π/15

=π

π/4

=π

cos 2x dx = π/2 0

(4, 16)

y = cos x

y = 4x

y = sin x x

y = x2

x 4

π 2x ln 3 e = 4π 2 0

ln 3

e2x dx = 0

14. V = π 0

π (1 − e−4 ) 4

e−4x dx =

13. V = π

-1

us uf

286

-1

16. V = 0

1 e6x π π 6x π dx = ln(1 + e ) = (ln(1 + e6 ) − ln 2) 6x 1+e 6 6 0

y 2/3 dy = 3π/5

17. V = π

2 1 π −1 15. V = π dx = tan (x/2) = π 2 /4 2 2 −2 4 + x −2

18. V = π −1

(1 − y 2 )2 dy

=π −1

(1 − 2y 2 + y 4 )dy = 16π/15 y 1

y1/3

x = 1 – y2

x -1

-1

uh a

-1

x= y = x3

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Exercise Set 7.2

(1 + y)dy = 8π

[22 − (y + 1)]dy

20. V = π

−1

(3 − y)dy = 9π/2

x = √y + 1 y = x2 – 1

y x = √1 + y x 2

(2, 3)

3π/4

csc2 y dy = 2π

(y − y 4 )dy = 3π/10

22. V = π

π/4

21. V = π

x = y2

x = csc y

(1, 1) x = √y

-1

2 2

23. V = π −1

[(y + 2) − y ]dy = 72π/5

x = y2

(4, 2) x= y+2 x

24. V = π

-1

-2

−1

(2 + y 2 )2 − (1 − y 2 )2 dy

(3 + 6y 2 )dy = 10π

=π −1

y x = 2 + y2 x = 1 – y2 1 x

(1, –1)

πe2y dy =

25. V =

uh a

27. V = π

π 2 e −1 2

-1

26. V = 0

π dy = π tan−1 2 1 + y2

b2 2 (a − x2 )dx = 4πab2 /3 a2

y b

y = ba √a2 – x 2 x

–a

−a

=π

us uf

19. V = π

287

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288

Chapter 7

1 dx = π(1/b − 1/2); 2 x b π(1/b − 1/2) = 3, b = 2π/(π + 6) (x + 1)dx

30. V = π

−1

= π/2 + π/2 = π

y = √x

y (1, √2) 1

y=6–x

y = √2x

(9 − y 2 )2 dy 0

(81 − 18y 2 + y 4 )dy

=π

an ss

33. V = π

√ (6 x − x)dx

=π

y=3 y = √x

x 9

√ [( x + 1)2 − (x + 1)2 ]dx

y x=y x = y2

√ (2 x − x − x2 )dx = π/2

x 1

y = -1

= 135π/2

x = y2

[(y + 1)2 − (y 2 + 1)2 ]dy

34. V = π

x=y

(2y − y 2 − y 4 )dy = 7π/15

=π

x)2 ]dx

= 648π/5

√

=π

[32 − (3 −

32. V = π

31. V = π

-1

= 8π + 8π/3 = 32π/3

[(x + 1) − 2x]dx

+π

y = √x + 1

(6 − x)2 dx

0 1

x dx + π

29. V = π

us uf

28. V = π

x 1 x = y2 x = –1

uh a

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Exercise Set 7.2

35. A(x) = π(x2 /4)2 = πx4 /16, 20 V = (πx4 /16)dx = 40, 000π ft3

(x − x4 )dx = 3π/10

36. V = π

(x − x2 )2 dx

38. A(x) =

1 2

(x − 2x + x )dx = 1/30

V =

Square

1 π 2

2 1√ 1 x = πx, 2 8

1 πx dx = π 8 y

y y = x (1, 1)

y = √x

y = x2 x

39. On the upper half of the circle, y =

√

us uf

37. V =

1 − x2 , so:

(a) A(x) is the area of a semicircle of radius y, so 1 π 1 2 2 2 A(x) = πy /2 = π(1 − x )/2; V = (1 − x ) dx = π (1 − x2 ) dx = 2π/3 2 −1 0 y

y = √1 – x2

-1

1 x

(b) A(x) is the area of a square of side 2y, so 1 1 A(x) = 4y 2 = 4(1 − x2 ); V = 4 (1 − x2 ) dx = 8 (1 − x2 ) dx = 16/3 −1

-1

y = √1 – x 2

1 x

(c) A(x) is the area of an equilateral triangle with sides 2y, so √ √ √ 3 A(x) = (2y)2 = 3y 2 = 3(1 − x2 ); 4 1√ √ 1 √ V = 3(1 − x2 ) dx = 2 3 (1 − x2 ) dx = 4 3/3

uh a

−1

-1

2y y = √1 – x2

289

1 x

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Chapter 7

40. By similar triangles, R/r = y/h so

R = ry/h and A(y) = πr2 y 2 /h2 . h 2 2 V = (πr /h ) y 2 dy = πr2 h/3

us uf

290

41. The two curves cross at x = b ≈ 1.403288534, so b π/2 16 2 V =π (sin16 x − (2x/π)2 ) dx ≈ 0.710172176. ((2x/π) − sin x) dx + π b

π/4

[(π 2 sin x cos3 x)2 − (4x2 )2 ] dx =

V =π

1 5 17 6 π + π 48 2560

(1 − (ln y)2 ) dy = π

43. V = π 1

tan 1

π[x2 − x2 tan−1 x] dx =

π [tan2 1 − ln(1 + tan2 1)] 6

44. V =

(r2 − y 2 ) dy = π(rh2 − h3 /3) =

45. (a) V = π r−h

(b) By the Pythagorean Theorem,

1 2 πh (3r − h) 3

42. Note that π 2 sin x cos3 x = 4x2 for x = π/4. From the graph it is apparent that this is the ﬁrst positive solution, thus the curves don’t cross on (0, π/4) and

h r

46. Find the volume generated by revolving the shaded region about the y-axis. −10+h π (100 − y 2 )dy = h2 (30 − h) V =π 3 −10

y h – 10 h -10

x = √100 – y 2

Find dh/dt when h = 5 given that dV /dt = 1/2.

x2 + y2 = r2 x

r2 = (r − h)2 + ρ2 , 2hr = h2 + ρ2 ; from Part (a), πh πh 3 2 2 2 2 (3hr − h ) = V = (h + ρ ) − h ) 3 2 3 1 = πh(h2 + 3ρ2 ) 6

π dV dh π (30h2 − h3 ), = (60h − 3h2 ) , 3 dt 3 dt π 1 dh dh = (300 − 75) , = 1/(150π) ft/min 3 dt dt 2

uh a

V =

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Exercise Set 7.2

291

5 = 0.5; {y0 , y1 , · · · , y10 } = {0, 2.00, 2.45, 2.45, 2.00, 1.46, 1.26, 1.25, 1.25, 1.25, 1.25}; 10 9 y i 2 ∆x ≈ 11.157; left = π 2 i=0 right = π

10 y i 2

i=1

∆x ≈ 11.771; V ≈ average = 11.464 cm3

√ 48. If x = r/2 then from y 2 = r2 − x2 we get y = ± 3r/2 √ √ as limits of integration; for − 3 ≤ y ≤ 3,

√3r 2

A(y) = π[(r2 − y 2 ) − r2 /4] = π(3r2 /4 − y 2 ), thus

x = √r 2 – y 2 x

√ 3r/2

(3r2 /4 − y 2 )dy

√ − 3r/2 √3r/2

(3r2 /4 − y 2 )dy =

= 2π

√

– √3r 2

V =π

us uf

47. (b) ∆x =

3πr3 /2

49. (a)

(b)

h –4

-2

h–4 h

-4

2≤h≤4

-4 0≤h<2

If the cherry is partially submerged then 0 ≤ h < 2 as shown in Figure (a); if it is totally submerged then 2 ≤ h ≤ 4 as shown in Figure (b). The radius of the glass is 4 cm and that of the cherry is 1 cm so points on the sections shown in the ﬁgures satisfy the equations x2 + y 2 = 16 and x2 + (y + 3)2 = 1. We will ﬁnd the volumes of the solids that are generated when the shaded regions are revolved about the y-axis. For 0 ≤ h < 2,

ad −4

for 2 ≤ h ≤ 4,

−2

V =π

−4

uh a M

[(16 − y 2 ) − (1 − (y + 3)2 )]dy = 6π

[(16 − y 2 ) − (1 − (y + 3)2 )]dy + π

−2

= 6π

h−4

V =π

h−4

(y + 4)dy + π −4

−2

h−4

(y + 4)dy = 3πh2 ; −4

h−4

−2

(16 − y 2 )dy

1 (16 − y 2 )dy = 12π + π(12h2 − h3 − 40) 3

1 π(12h2 − h3 − 4) 3  2  3πh V =  1 π(12h2 − h3 − 4) 3

if 0 ≤ h < 2 if 2 ≤ h ≤ 4

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Chapter 7

x = h Âą r2 â&#x2C6;&#x2019; y2 , r (h + r2 â&#x2C6;&#x2019; y 2 )2 â&#x2C6;&#x2019; (h â&#x2C6;&#x2019; r2 â&#x2C6;&#x2019; y 2 )2 dy V =Ď&#x20AC; â&#x2C6;&#x2019;r

= 4Ď&#x20AC;h = 4Ď&#x20AC;h

r2 â&#x2C6;&#x2019; y 2 dy

â&#x2C6;&#x2019;r

(x â&#x20AC;&#x201C; h 2) + y 2 = r 2

1 2 Ď&#x20AC;r 2

= 2Ď&#x20AC; 2 r2 h

1 1 1 hx = x2 tan Î¸ = (r2 â&#x2C6;&#x2019; y 2 ) tan Î¸ 2 2 2

because x = r â&#x2C6;&#x2019; y , r 1 V = tan Î¸ (r2 â&#x2C6;&#x2019; y 2 )dy 2 â&#x2C6;&#x2019;r r 2 = tan Î¸ (r2 â&#x2C6;&#x2019; y 2 )dy = r3 tan Î¸ 3 0 2

u x

52. A(x) = (x tan Î¸)(2 r2 â&#x2C6;&#x2019; x2 ) = 2(tan Î¸)x r2 â&#x2C6;&#x2019; x2 , r V = 2 tan Î¸ x r2 â&#x2C6;&#x2019; x2 dx

53. Each cross section perpendicular to the y-axis is a square so

A(y) = x2 = r2 â&#x2C6;&#x2019; y 2 , r 1 (r2 â&#x2C6;&#x2019; y 2 )dy V = 8 0

2 = r3 tan Î¸ 3

V = 8(2r3 /3) = 16r3 /3

x tan u y

x â&#x20AC;&#x201C; x2

â&#x2C6;&#x161;r 2

51. tan Î¸ = h/x so h = x tan Î¸, A(y) =

50.

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us uf

292

y x = â&#x2C6;&#x161;r2 â&#x20AC;&#x201C; y2

r x

54. The regular cylinder of radius r and height h has the same circular cross sections as do those of the oblique clinder, so by Cavalieriâ&#x20AC;&#x2122;s Principle, they have the same volume: Ď&#x20AC;r2 h.

EXERCISE SET 7.3 2

2Ď&#x20AC;x(x2 )dx = 2Ď&#x20AC;

1. V =

2. V =

x3 dx = 15Ď&#x20AC;/2 1

2Ď&#x20AC;x( 4 â&#x2C6;&#x2019; x2 â&#x2C6;&#x2019; x)dx = 2Ď&#x20AC;

uh a

â&#x2C6;&#x161; 2

2Ď&#x20AC;y(2y â&#x2C6;&#x2019; 2y 2 )dy = 4Ď&#x20AC;

3. V =

â&#x2C6;&#x161; 8Ď&#x20AC; (x 4 â&#x2C6;&#x2019; x2 â&#x2C6;&#x2019; x2 )dx = (2 â&#x2C6;&#x2019; 2) 3

(y 2 â&#x2C6;&#x2019; y 3 )dy = Ď&#x20AC;/3 0

â&#x2C6;&#x161; 2

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Exercise Set 7.3

2πy[y − (y 2 − 2)]dy = 2π 0

5. V =

2π(x)(x )dx 0

6. V = 4

x4 dx = 2π/5

= 2π

√ 2πx( x)dx

x3/2 dx = 844π/5

= 2π

y = √x

y = x3 2

1 x -1

us uf

(y 2 − y 3 + 2y)dy = 16π/3

4. V =

-9

-4

7. V =

2πx(1/x)dx = 2π

dx = 4π

y = cos (x2)

1 x

2πx[(2x − 1) − (−2x + 3)]dx 1

2 2

(x − x)dx = 20π/3

= 8π 1

(2, 3)

2πx(2x − x2 )dx

10. V = 0

9. V =

√p 2

-1

√ 2πx cos(x2 )dx = π/ 2

-3

√ π/2

8. V =

-1

293

(2x2 − x3 )dx =

= 2π 0

8 π 3

y = 2x – x 2

(1, 1)

x 2

(2, –1)

x dx +1 0 1 = π ln(x2 + 1) = π ln 2

1 x2 + 1

x -1

uh a

11. V = 2π

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Chapter 7

√ 3

12. V =

2πxe

dx = πe

√3

= π(e3 − e)

y 20 y = ex

us uf

294

x -√3 -1

2πy 3 dy = π/2

14. V =

13. V =

√3

y 3 2

x = y2

y 2 dy = 76π/3

2πy(2y)dy = 4π

x = 2y

2πy(1 − 0

√

y)dy

2πy(5 − y − 4/y)dy

16. V = 1

(y − y 3/2 )dy = π/5

= 2π 0

y = √x x

x sin xdx = 2π 2

17. V = 2π

(5y − y 2 − 4)dy = 9π y (1, 4)

x = 5–y (4, 1) x = 4/y

π/2

x cos xdx = π 2 − 2π

18. V = 2π 0

19. (a) V =

= 2π

15. V =

2πx(x3 − 3x2 + 2x)dx = 7π/30

uh a

(b) much easier; the method of slicing would require that x be expressed in terms of y.

x 1

-1

y y = x3 – 3x2 + 2x

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Exercise Set 7.3

2Ï&#x20AC;(x + 1)(1/x3 )dx

= 2Ï&#x20AC;

y x+1

(xâ&#x2C6;&#x2019;2 + xâ&#x2C6;&#x2019;3 )dx = 7Ï&#x20AC;/4

y = 1/x 3

x 1x 2

-1

2Ï&#x20AC;(1 â&#x2C6;&#x2019; y)y 1/3 dy

21. V =

(y 1/3 â&#x2C6;&#x2019; y 4/3 )dy = 9Ï&#x20AC;/14

= 2Ï&#x20AC; 0

x = y1/3

1â&#x20AC;&#x201C;y

2Ï&#x20AC;x[f (x) â&#x2C6;&#x2019; g(x)]dx

2Ï&#x20AC;y[f (y) â&#x2C6;&#x2019; g(y)]dy

(b) c

h (r â&#x2C6;&#x2019; y) is an equation of the line r through (0, r) and (h, 0) so

r h (r â&#x2C6;&#x2019; y) dy V = 2Ï&#x20AC;y r 0 2Ï&#x20AC;h r = (ry â&#x2C6;&#x2019; y 2 )dy = Ï&#x20AC;r2 h/3 r 0

22. (a)

23. x =

(0, r)

(h, 0)

k/4

24. V =

â&#x2C6;&#x161; 2Ï&#x20AC;(k/2 â&#x2C6;&#x2019; x)2 kxdx

â&#x2C6;&#x161; = 2Ï&#x20AC; k

k/4

(kx1/2 â&#x2C6;&#x2019; 2x3/2 )dx = 7Ï&#x20AC;k 3 /60

k/2 â&#x20AC;&#x201C; x

y = â&#x2C6;&#x161;kx x

a 2Ï&#x20AC;x(2 r2 â&#x2C6;&#x2019; x2 )dx = 4Ï&#x20AC; x(r2 â&#x2C6;&#x2019; x2 )1/2 dx 0 0 a 4Ï&#x20AC; 3 4Ï&#x20AC; r â&#x2C6;&#x2019; (r2 â&#x2C6;&#x2019; a2 )3/2 = â&#x2C6;&#x2019; (r2 â&#x2C6;&#x2019; x2 )3/2 = 3 3 0

x = k/2 x = k/4

y = â&#x2C6;&#x161;r 2 â&#x20AC;&#x201C; x2 x

a y = â&#x20AC;&#x201C;â&#x2C6;&#x161;r 2 â&#x20AC;&#x201C; x2

uh a

y = â&#x20AC;&#x201C;â&#x2C6;&#x161;kx

us uf

20. V =

25. V =

295

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Chapter 7

â&#x2C6;&#x2019;a

2Ï&#x20AC;(b â&#x2C6;&#x2019; x)(2 a2 â&#x2C6;&#x2019; x2 )dx

= 4Ï&#x20AC;b â&#x2C6;&#x2019;a

a2 â&#x2C6;&#x2019; x2 dx â&#x2C6;&#x2019; 4Ï&#x20AC;

â&#x2C6;&#x2019;a

x a2 â&#x2C6;&#x2019; x2 dx

b-x

â&#x2C6;&#x161;a2 â&#x20AC;&#x201C; x2 x â&#x20AC;&#x201C;a

= 4Ï&#x20AC;b Â· (area of a semicircle of radius a) â&#x2C6;&#x2019; 4Ï&#x20AC;(0)

â&#x20AC;&#x201C;â&#x2C6;&#x161;a2 â&#x20AC;&#x201C; x2

= 2Ï&#x20AC; 2 a2 b b

1/2

x=b

1 dx = Ï&#x20AC;(2 â&#x2C6;&#x2019; 1/b), Vy = 2Ï&#x20AC; x2

dx = Ï&#x20AC;(2b â&#x2C6;&#x2019; 1); 1/2

27. Vx = Ï&#x20AC;

Vx = Vy if 2 â&#x2C6;&#x2019; 1/b = 2b â&#x2C6;&#x2019; 1, 2b2 â&#x2C6;&#x2019; 3b + 1 = 0, solve to get b = 1/2 (reject) or b = 1. b x Ï&#x20AC; â&#x2C6;&#x2019;1 2 â&#x2C6;&#x2019;1 2 ) â&#x2C6;&#x2019; dx = Ï&#x20AC; tan (x ) = Ï&#x20AC; tan (b 4 4 1 1+x 1 Ï&#x20AC; Ï&#x20AC; 1 â&#x2C6;&#x2019; = Ï&#x20AC;2 (b) lim V = Ï&#x20AC; bâ&#x2020;&#x2019;+â&#x2C6;&#x17E; 2 4 4

28. (a) V = 2Ï&#x20AC;

EXERCISE SET 7.4 â&#x2C6;&#x161;

1 + 4dx =

â&#x2C6;&#x161;

â&#x2C6;&#x161; â&#x2C6;&#x161; 1 + 1/4 dy = 2 5/2 = 5

dx dy = 1, = 5, L = dt dt

12 + 52 dt =

â&#x2C6;&#x161;

dx 1 (b) = ,L = dy 2

dy = 2, L = dx

1. (a)

9 1/2 81 x , 1 + [f (x)]2 = 1 + x, 2 4 3/2 1 1 â&#x2C6;&#x161; 81 8 1+ x L= 1 + 81x/4 dx = = (85 85 â&#x2C6;&#x2019; 8)/243 243 4 0

3. f (x) =

4. g (y) = y(y 2 + 2)1/2 , 1 + [g (y)]2 = 1 + y 2 (y 2 + 2) = y 4 + 2y 2 + 1 = (y 2 + 1)2 , 1 (y 2 + 1)2 dy = (y 2 + 1)dy = 4/3 0

26. V =

us uf

296

uh a

2 dy dy 4 9x2/3 + 4 2 â&#x2C6;&#x2019;1/3 5. , 1+ = 1 + xâ&#x2C6;&#x2019;2/3 = , = x dx 3 dx 9 9x2/3 40 8 â&#x2C6;&#x161; 2/3 9x + 4 1 L= dx = u1/2 du, u = 9x2/3 + 4 1/3 18 3x 1 13 40 â&#x2C6;&#x161; â&#x2C6;&#x161; â&#x2C6;&#x161; â&#x2C6;&#x161; 1 3/2 1 1 = = u (40 40 â&#x2C6;&#x2019; 13 13) = (80 10 â&#x2C6;&#x2019; 13 13) 27 27 27 13

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Exercise Set 7.4

297

or (alternate solution) 2 4 + 9y dx 9 , =1+ y = dy 4 4 40 â&#x2C6;&#x161; â&#x2C6;&#x161; 1 4 1 1 L= (80 10 â&#x2C6;&#x2019; 13 13) 4 + 9y dy = u1/2 du = 2 1 18 13 27

3 dx = y 1/2 , 1 + dy 2

us uf

x = y 3/2 ,

2 1 6 1 1 3 1 6 1 1 3 â&#x2C6;&#x2019;3 2 â&#x2C6;&#x2019;6 â&#x2C6;&#x2019;6 â&#x2C6;&#x2019;3 x â&#x2C6;&#x2019; +x x + +x = x +x 6. f (x) = x â&#x2C6;&#x2019; x , 1 + [f (x)] = 1 + = , 16 2 16 2 4 4 2 3 3 1 3 1 3 â&#x2C6;&#x2019;3 â&#x2C6;&#x2019;3 dx = 595/144 dx = L= x +x x +x 4 4 2 2

1 3 1 y + 2y â&#x2C6;&#x2019;1 , g (y) = y 2 â&#x2C6;&#x2019; 2y â&#x2C6;&#x2019;2 , 8 24 2 1 4 1 1 2 1 4 1 â&#x2C6;&#x2019;2 â&#x2C6;&#x2019;4 â&#x2C6;&#x2019;4 2 y + 2y y â&#x2C6;&#x2019; + 4y y + + 4y = = , 1 + [g (y)] = 1 + 2 2 8 64 64 4 1 2 y + 2y â&#x2C6;&#x2019;2 dy = 17/6 L= 8 2

1 6 1 1 â&#x2C6;&#x2019;6 y â&#x2C6;&#x2019; + y 4 2 4

2 1 3 1 â&#x2C6;&#x2019;3 y + y , 2 2

1 1 8. g (y) = y 3 â&#x2C6;&#x2019; y â&#x2C6;&#x2019;3 , 1 + [g (y)]2 = 1 + 2 2 4 1 3 1 â&#x2C6;&#x2019;3 L= dy = 2055/64 y + y 2 2 1

7. x = g(y) =

9. (dx/dt)2 + (dy/dt)2 = (t2 )2 + (t)2 = t2 (t2 + 1), L =

â&#x2C6;&#x161; t(t2 + 1)1/2 dt = (2 2 â&#x2C6;&#x2019; 1)/3

10. (dx/dt)2 + (dy/dt)2 = [2(1 + t)]2 + [3(1 + t)2 ]2 = (1 + t)2 [4 + 9(1 + t)2 ], 1 â&#x2C6;&#x161; â&#x2C6;&#x161; L= (1 + t)[4 + 9(1 + t)2 ]1/2 dt = (80 10 â&#x2C6;&#x2019; 13 13)/27

Ď&#x20AC;/2

11. (dx/dt) + (dy/dt) = (â&#x2C6;&#x2019;2 sin 2t) + (2 cos 2t) = 4, L =

2 dt = Ď&#x20AC;

12. (dx/dt)2 + (dy/dt)2 = (â&#x2C6;&#x2019; sin t + sin t + t cos t)2 + (cos t â&#x2C6;&#x2019; cos t + t sin t)2 = t2 , Ď&#x20AC; t dt = Ď&#x20AC; 2 /2 L=

13. (dx/dt)2 + (dy/dt)2 = [et (cos t â&#x2C6;&#x2019; sin t)]2 + [et (cos t + sin t)]2 = 2e2t , Ď&#x20AC;/2 â&#x2C6;&#x161; â&#x2C6;&#x161; L= 2et dt = 2(eĎ&#x20AC;/2 â&#x2C6;&#x2019; 1)

uh a

2et dt = 2(e4 â&#x2C6;&#x2019; e)

14. (dx/dt)2 + (dy/dt)2 = (2et cos t)2 + (â&#x2C6;&#x2019;2et sin t)2 = 4e2t , L = 1

â&#x2C6;&#x161; sec x tan x = tan x, 1 + (y )2 = 1 + tan2 x = sec x when 0 < x < Ď&#x20AC;/4, so sec x Ď&#x20AC;/4 â&#x2C6;&#x161; sec x dx = ln(1 + 2) L=

15. dy/dx =

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298

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Chapter 7

â&#x2C6;&#x161; cos x = cot x, 1 + (y )2 = 1 + cot2 x = csc x when Ď&#x20AC;/4 < x < Ď&#x20AC;/2, so sin x â&#x2C6;&#x161; Ď&#x20AC;/2 â&#x2C6;&#x161; â&#x2C6;&#x161; 2â&#x2C6;&#x2019;1 â&#x2C6;&#x161; csc x dx = â&#x2C6;&#x2019; ln( 2 â&#x2C6;&#x2019; 1) = â&#x2C6;&#x2019; ln â&#x2C6;&#x161; ( 2 + 1) = ln(1 + 2) L= 2+1 Ď&#x20AC;/4

18. (a) Use the interval 0 â&#x2030;¤ Ď&#x2020; < 2Ď&#x20AC;. (b) (dx/dĎ&#x2020;)2 + (dy/dĎ&#x2020;)2 = (â&#x2C6;&#x2019;3a cos2 Ď&#x2020; sin Ď&#x2020;)2 + (3a sin2 Ď&#x2020; cos Ď&#x2020;)2

us uf

17. (a) (dx/dÎ¸)2 + (dy/dÎ¸)2 = (a(1 â&#x2C6;&#x2019; cos Î¸))2 + (a sin Î¸)2 = a2 (2 â&#x2C6;&#x2019; 2 cos Î¸), so 2Ď&#x20AC; 2Ď&#x20AC; 2 2 L= (dx/dÎ¸) + (dy/dÎ¸) dÎ¸ = a 2(1 â&#x2C6;&#x2019; cos Î¸) dÎ¸

16. dy/dx =

= 9a2 cos2 Ď&#x2020; sin2 Ď&#x2020;(cos2 Ď&#x2020; + sin2 Ď&#x2020;) = (9a2 /4) sin2 2Ď&#x2020;, so Ď&#x20AC;/2 2Ď&#x20AC; Ď&#x20AC;/2 L = (3a/2) sin 2Ď&#x2020; dĎ&#x2020; = â&#x2C6;&#x2019;3a cos 2Ď&#x2020; = 6a | sin 2Ď&#x2020;| dĎ&#x2020; = 6a 0

19. (a)

(b) dy/dx does not exist at x = 0.

(8, 4) (-1, 1)

x = g(y) = y 3/2 , g (y) = 1

1 + 9y/4 dy

+ 0

8 = 27

(portion for â&#x2C6;&#x2019; 1 â&#x2030;¤ x â&#x2030;¤ 0)

1 + 9y/4 dy

3 1/2 y , 2

(portion for 0 â&#x2030;¤ x â&#x2030;¤ 8)

(c)

â&#x2C6;&#x161; â&#x2C6;&#x161; â&#x2C6;&#x161; 13 â&#x2C6;&#x161; 8 13 â&#x2C6;&#x2019; 1 + (10 10 â&#x2C6;&#x2019; 1) = (13 13 + 80 10 â&#x2C6;&#x2019; 16)/27 8 27

1 + 4x2 dx â&#x2030;&#x2C6; 4.645975301

21. L =

20. For (4), express the curve y = f (x) in the parametric form x = t, y = f (t) so dx/dt = 1 and dy/dt = f (t) = f (x) = dy/dx. For (5), express x = g(y) as x = g(t), y = t so dx/dt = g (t) = g (y) = dx/dy and dy/dt = 1. 22. L =

Ď&#x20AC;

1 + cos2 y dy â&#x2030;&#x2C6; 3.820197789

23. Numerical integration yields: in Exercise 21, L â&#x2030;&#x2C6; 4.646783762; in Exercise 22, L â&#x2030;&#x2C6; 3.820197788.

uh a

24. 0 â&#x2030;¤ m â&#x2030;¤ f (x) â&#x2030;¤ M , so m2 â&#x2030;¤ [f (x)]2 â&#x2030;¤ M 2 , and 1 + m2 â&#x2030;¤ 1 + [f (x)]2 â&#x2030;¤ 1 + M 2 ; thus â&#x2C6;&#x161; â&#x2C6;&#x161; 1 + m2 â&#x2030;¤ 1 + [f (x)]2 â&#x2030;¤ 1 + M 2 , b b b 2 2 1 + m dx â&#x2030;¤ 1 + [f (x)] dx â&#x2030;¤ 1 + M 2 dx, and a a a (b â&#x2C6;&#x2019; a) 1 + m2 â&#x2030;¤ L â&#x2030;¤ (b â&#x2C6;&#x2019; a) 1 + M 2

â&#x2C6;&#x161; 25. f (x) = cos x, 2/2 â&#x2030;¤ cos x â&#x2030;¤ 1 for 0 â&#x2030;¤ x â&#x2030;¤ Ď&#x20AC;/4 so â&#x2C6;&#x161; Ď&#x20AC; Ď&#x20AC;â&#x2C6;&#x161; 3/2 â&#x2030;¤ L â&#x2030;¤ 2. (Ď&#x20AC;/4) 1 + 1/2 â&#x2030;¤ L â&#x2030;¤ (Ď&#x20AC;/4) 1 + 1, 4 4

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Exercise Set 7.5

299

26. (dx/dt)2 + (dy/dt)2 = (â&#x2C6;&#x2019;a sin t)2 + (b cos t)2 = a2 sin2 t + b2 cos2 t

27. (a) (dx/dt)2 + (dy/dt)2 = 4 sin2 t + cos2 t = 4 sin2 t + (1 â&#x2C6;&#x2019; sin2 t) = 1 + 3 sin2 t, 2Ď&#x20AC; Ď&#x20AC;/2 L= 1 + 3 sin2 t dt = 4 1 + 3 sin2 t dt 0

(b) 9.69

4.8

1 + 3 sin2 t dt â&#x2030;&#x2C6; 5.16 cm

4.6

1 + (2.09 â&#x2C6;&#x2019; 0.82x)2 dx â&#x2030;&#x2C6; 6.65 m

Ď&#x20AC;

29. L =

1.5

28. The distance is

us uf

= a2 (1 â&#x2C6;&#x2019; cos2 t) + b2 cos2 t = a2 â&#x2C6;&#x2019; (a2 â&#x2C6;&#x2019; b2 ) cos2 t

a2 â&#x2C6;&#x2019; b2 2 2 =a 1â&#x2C6;&#x2019; cos t = a2 [1 â&#x2C6;&#x2019; k 2 cos2 t], a2 2Ď&#x20AC; Ď&#x20AC;/2 L= a 1 â&#x2C6;&#x2019; k 2 cos2 t dt = 4a 1 â&#x2C6;&#x2019; k 2 cos2 t dt

1 + (k cos x)2 dx

1.84

1.83

1.832

3.8202

5.2704

5.0135

4.9977

5.0008

1. S =

EXERCISE SET 7.5

Experimentation yields the values in the table, which by the Intermediate-Value Theorem show that the true solution k to L = 5 lies between k = 1.83 and k = 1.832, so k = 1.83 to two decimal places.

â&#x2C6;&#x161; â&#x2C6;&#x161; 2Ď&#x20AC;(7x) 1 + 49dx = 70Ď&#x20AC; 2

â&#x2C6;&#x161; x dx = 35Ď&#x20AC; 2

1 1 2. f (x) = â&#x2C6;&#x161; , 1 + [f (x)]2 = 1 + 4x 2 x 4 4 â&#x2C6;&#x161; â&#x2C6;&#x161; â&#x2C6;&#x161; 1 dx = 2Ď&#x20AC; 2Ď&#x20AC; x 1 + x + 1/4dx = Ď&#x20AC;(17 17 â&#x2C6;&#x2019; 5 5)/6 S= 4x 1 1 â&#x2C6;&#x161; 3. f (x) = â&#x2C6;&#x2019;x/ 4 â&#x2C6;&#x2019; x2 , 1 + [f (x)]2 = 1 +

x2 4 = , 2 4â&#x2C6;&#x2019;x 4 â&#x2C6;&#x2019; x2 1 2 2 2Ď&#x20AC; 4 â&#x2C6;&#x2019; x (2/ 4 â&#x2C6;&#x2019; x )dx = 4Ď&#x20AC; dx = 8Ď&#x20AC;

â&#x2C6;&#x2019;1

uh a

â&#x2C6;&#x2019;1

4. y = f (x) = x3 for 1 â&#x2030;¤ x â&#x2030;¤ 2, f (x) = 3x2 , 2 2 â&#x2C6;&#x161; â&#x2C6;&#x161; Ď&#x20AC; 3 4 3/2 4 S= 2Ď&#x20AC;x 1 + 9x dx = = 5Ď&#x20AC;(29 145 â&#x2C6;&#x2019; 2 10)/27 (1 + 9x ) 27 1 1

5. S = 0

â&#x2C6;&#x161; â&#x2C6;&#x161; 2 â&#x2C6;&#x161; 2Ď&#x20AC;(9y + 1) 82dy = 2Ď&#x20AC; 82 (9y + 1)dy = 40Ď&#x20AC; 82 0

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300

Chapter 7

6. g (y) = 3y 2 , S =

2Ï&#x20AC;y 3

â&#x2C6;&#x161; 1 + 9y 4 dy = Ï&#x20AC;(10 10 â&#x2C6;&#x2019; 1)/27

9 â&#x2C6;&#x2019; y 2 , 1 + [g (y)]2 =

9 ,S= 9 â&#x2C6;&#x2019; y2

2Ï&#x20AC;

â&#x2C6;&#x2019;2

9 â&#x2C6;&#x2019; y2 Â·

3 9 â&#x2C6;&#x2019; y2

dy = 6Ï&#x20AC;

2â&#x2C6;&#x2019;y , 1â&#x2C6;&#x2019;y â&#x2C6;&#x161; 0 0 â&#x2C6;&#x161; â&#x2C6;&#x161; 2â&#x2C6;&#x2019;y dy = 4Ï&#x20AC; 2Ï&#x20AC;(2 1 â&#x2C6;&#x2019; y) â&#x2C6;&#x161; 2 â&#x2C6;&#x2019; y dy = 8Ï&#x20AC;(3 3 â&#x2C6;&#x2019; 2 2)/3 S= 1â&#x2C6;&#x2019;y â&#x2C6;&#x2019;1 â&#x2C6;&#x2019;1

â&#x2C6;&#x2019;2

8. g (y) = â&#x2C6;&#x2019;(1 â&#x2C6;&#x2019; y)â&#x2C6;&#x2019;1/2 , 1 + [g (y)]2 =

dy = 24Ï&#x20AC;

us uf

7. g (y) = â&#x2C6;&#x2019;y/

2 1 1 1 1 â&#x2C6;&#x2019;1/2 1 1/2 1 â&#x2C6;&#x2019;1/2 1 1/2 x + x , â&#x2C6;&#x2019; x , 1 + [f (x)]2 = 1 + xâ&#x2C6;&#x2019;1 â&#x2C6;&#x2019; + x = x 2 2 4 2 4 2 2 3 Ï&#x20AC; 3 1 â&#x2C6;&#x2019;1/2 1 1/2 1 dx = + x (3 + 2x â&#x2C6;&#x2019; x2 )dx = 16Ï&#x20AC;/9 2Ï&#x20AC; x1/2 â&#x2C6;&#x2019; x3/2 S= x 2 3 2 3 1 1

9. f (x) =

2 1 â&#x2C6;&#x2019;4 1 â&#x2C6;&#x2019;2 1 1 â&#x2C6;&#x2019;2 2 4 2 = x + x , 10. f (x) = x â&#x2C6;&#x2019; x , 1 + [f (x)] = 1 + x â&#x2C6;&#x2019; + x 4 2 16 4 2 2 1 â&#x2C6;&#x2019;3 1 5 1 1 3 1 â&#x2C6;&#x2019;1 1 â&#x2C6;&#x2019;2 2 x + x x + x+ x S= x + x dx = 2Ï&#x20AC; dx = 515Ï&#x20AC;/64 2Ï&#x20AC; 3 4 4 3 3 16 1 1

1 4 1 â&#x2C6;&#x2019;2 1 y + y , g (y) = y 3 â&#x2C6;&#x2019; y â&#x2C6;&#x2019;3 , 8 4 4 2 1 1 1 1 + [g (y)]2 = 1 + y 6 â&#x2C6;&#x2019; + y â&#x2C6;&#x2019;6 = y 3 + y â&#x2C6;&#x2019;3 , 2 16 4 2 2 Ï&#x20AC; 1 4 1 â&#x2C6;&#x2019;2 1 S= y 3 + y â&#x2C6;&#x2019;3 dy = 2Ï&#x20AC; (8y 7 + 6y + y â&#x2C6;&#x2019;5 )dy = 16,911Ï&#x20AC;/1024 y + y 4 8 4 16 1 1

11. x = g(y) =

â&#x2C6;&#x161;

1 65 â&#x2C6;&#x2019; 4y , , 1 + [g (y)]2 = 16 â&#x2C6;&#x2019; y; g (y) = â&#x2C6;&#x2019; â&#x2C6;&#x161; 4(16 â&#x2C6;&#x2019; y) 2 16 â&#x2C6;&#x2019; y 15 15 â&#x2C6;&#x161; â&#x2C6;&#x161; Ï&#x20AC; 65 â&#x2C6;&#x2019; 4y S= 2Ï&#x20AC; 16 â&#x2C6;&#x2019; y 65 â&#x2C6;&#x2019; 4y dy = (65 65 â&#x2C6;&#x2019; 5 5) dy = Ï&#x20AC; 4(16 â&#x2C6;&#x2019; y) 6 0 0

12. x = g(y) =

13. f (x) = cos x, 1 + [f (x)] = 1 + cos x, S =

Ï&#x20AC;

â&#x2C6;&#x161; â&#x2C6;&#x161; 2Ï&#x20AC; sin x 1 + cos2 x dx = 2Ï&#x20AC;( 2 + ln( 2 + 1))

14. x = g(y) = tan y, g (y) = sec2 y, 1 + [g (y)]2 = 1 + sec4 y; Ï&#x20AC;/4 S= 2Ï&#x20AC; tan y 1 + sec4 y dy â&#x2030;&#x2C6; 3.84 x

uh a

15. f (x) = e ,

2Ï&#x20AC;ex

1 + [f (x)] = 1 + e , S =

1 + e2x dx â&#x2030;&#x2C6; 22.94

16. x = g(y) = ln y, g (y) = 1/y, 1 + [g (y)]2 = 1 + 1/y 2 ; S =

2Ï&#x20AC;

1 + 1/y 2 ln y dy â&#x2030;&#x2C6; 7.05

17. Revolve the line segment joining the points (0, 0) and (h, r) about the x-axis. An equation of the line segment is y = (r/h)x for 0 â&#x2030;¤ x â&#x2030;¤ h so h h 2Ï&#x20AC;r S= 2Ï&#x20AC;(r/h)x 1 + r2 /h2 dx = 2 r2 + h2 x dx = Ï&#x20AC;r r2 + h2 h 0 0

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Exercise Set 7.5

301

â&#x2C6;&#x161; â&#x2C6;&#x161; 18. f (x) = r2 â&#x2C6;&#x2019; x2 , f (x) = â&#x2C6;&#x2019;x/ r2 â&#x2C6;&#x2019; x2 , 1 + [f (x)]2 = r2 /(r2 â&#x2C6;&#x2019; x2 ), r r 2 2 2 2 2Ď&#x20AC; r â&#x2C6;&#x2019; x (r/ r â&#x2C6;&#x2019; x )dx = 2Ď&#x20AC;r dx = 4Ď&#x20AC;r2 S=

â&#x2C6;&#x2019;r

us uf

â&#x2C6;&#x2019;r

r2 â&#x2C6;&#x2019; y 2 , g (y) = â&#x2C6;&#x2019;y/ r2 â&#x2C6;&#x2019; y 2 , 1 + [g (y)]2 = r2 /(r2 â&#x2C6;&#x2019; y 2 ), r r 2 2 2 2 2 2Ď&#x20AC; r â&#x2C6;&#x2019; y r /(r â&#x2C6;&#x2019; y ) dy = 2Ď&#x20AC;r dy = 2Ď&#x20AC;rh (a) S =

19. g(y) =

râ&#x2C6;&#x2019;h

(b) From Part (a), the surface area common to two polar caps of height h1 > h2 is 2Ď&#x20AC;rh1 â&#x2C6;&#x2019; 2Ď&#x20AC;rh2 = 2Ď&#x20AC;r(h1 â&#x2C6;&#x2019; h2 ).

20. For (4), express the curve y = f (x) in the parametric form x = t, y = f (t) so dx/dt = 1 and dy/dt = f (t) = f (x) = dy/dx. For (5), express x = g(y) as x = g(t), y = t so

dx/dt = g (t) = g (y) = dx/dy and dy/dt = 1.

21. x = 2t, y = 2, (x )2 + (y )2 = 4t2 + 4 4 4 â&#x2C6;&#x161; 8Ď&#x20AC; (17 17 â&#x2C6;&#x2019; 1) S = 2Ď&#x20AC; (2t) 4t2 + 4dt = 8Ď&#x20AC; t t2 + 1dt = 3 0 0

22. x = â&#x2C6;&#x2019;2 cos t sin t, y = 5 cos t, (x )2 + (y )2 = 4 cos2 t sin2 t + 25 cos2 t, Ď&#x20AC;/2 â&#x2C6;&#x161; Ď&#x20AC; S = 2Ď&#x20AC; 5 sin t 4 cos2 t sin2 t + 25 cos2 t dt = (145 29 â&#x2C6;&#x2019; 625) 6 0 23. x = 1, y = 4t, (x )2 + (y )2 = 1 + 16t2 , S = 2Ď&#x20AC;

1 + 16t2 dt =

â&#x2C6;&#x161; Ď&#x20AC; (17 17 â&#x2C6;&#x2019; 1) 24

24. x = â&#x2C6;&#x2019;2 sin t cos t, y = 2 sin t cos t, (x )2 + (y )2 = 8 sin2 t cos2 t Ď&#x20AC;/2 â&#x2C6;&#x161; Ď&#x20AC;/2 â&#x2C6;&#x161; S = 2Ď&#x20AC; cos2 t 8 sin2 t cos2 t dt = 4 2Ď&#x20AC; cos3 t sin t dt = 2Ď&#x20AC; 0

25. x = â&#x2C6;&#x2019;r sin t, y = r cos t, (x )2 + (y )2 = r2 , Ď&#x20AC; Ď&#x20AC; â&#x2C6;&#x161; r sin t r2 dt = 2Ď&#x20AC;r2 sin t dt = 4Ď&#x20AC;r2 S = 2Ď&#x20AC; 0

2 2 dy dx dy dx + = 2a2 (1 â&#x2C6;&#x2019; cos Ď&#x2020;) 26. = a(1 â&#x2C6;&#x2019; cos Ď&#x2020;), = a sin Ď&#x2020;, dĎ&#x2020; dĎ&#x2020; dĎ&#x2020; dĎ&#x2020; 2Ď&#x20AC; 2Ď&#x20AC; â&#x2C6;&#x161; 2 2 S = 2Ď&#x20AC; a(1 â&#x2C6;&#x2019; cos Ď&#x2020;) 2a (1 â&#x2C6;&#x2019; cos Ď&#x2020;) dĎ&#x2020; = 2 2Ď&#x20AC;a (1 â&#x2C6;&#x2019; cos Ď&#x2020;)3/2 dĎ&#x2020;, 0

uh a

â&#x2C6;&#x161; Ď&#x2020; Ď&#x2020; but 1 â&#x2C6;&#x2019; cos Ď&#x2020; = 2 sin2 so (1 â&#x2C6;&#x2019; cos Ď&#x2020;)3/2 = 2 2 sin3 for 0 â&#x2030;¤ Ď&#x2020; â&#x2030;¤ Ď&#x20AC; and, taking advantage of the 2 2 Ď&#x20AC; 3 Ď&#x2020; 2 sin dĎ&#x2020; = 64Ď&#x20AC;a2 /3. symmetry of the cycloid, S = 16Ď&#x20AC;a 2 0

27. (a) length of arc of sector = circumference of base of cone, 1 !Î¸ = 2Ď&#x20AC;r, Î¸ = 2Ď&#x20AC;r/!; S = area of sector = !2 (2Ď&#x20AC;r/!) = Ď&#x20AC;r! 2

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Chapter 7

l1 r1

us uf

(b) S = Ď&#x20AC;r2 !2 â&#x2C6;&#x2019; Ď&#x20AC;r1 !1 = Ď&#x20AC;r2 (!1 + !) â&#x2C6;&#x2019; Ď&#x20AC;r1 !1 = Ď&#x20AC;[(r2 â&#x2C6;&#x2019; r1 )!1 + r2 !]; Using similar triangles !2 /r2 = !1 /r1 , r1 !2 = r2 !1 , r1 (!1 + !) = r2 !1 , (r2 â&#x2C6;&#x2019; r1 )!1 = r1 ! so S = Ď&#x20AC; (r1 ! + r2 !) = Ď&#x20AC; (r1 + r2 ) !.

l2 l

28. S =

302

2Ď&#x20AC;[f (x) + k] 1 + [f (x)]2 dx

1 + [f (x)]2 dx â&#x2030;¤ S â&#x2030;¤ 2Ď&#x20AC;K

2Ď&#x20AC;k a

a b

1 + [f (x)]2 dx, 2Ď&#x20AC;kL â&#x2030;¤ S â&#x2030;¤ 2Ď&#x20AC;KL

1 + [f (x)]2 so 2Ď&#x20AC;f (x) â&#x2030;¤ 2Ď&#x20AC;f (x) 1 + [f (x)]2 , b b 2Ď&#x20AC;f (x)dx â&#x2030;¤ 2Ď&#x20AC;f (x) 1 + [f (x)]2 dx, 2Ď&#x20AC; f (x)dx â&#x2030;¤ S, 2Ď&#x20AC;A â&#x2030;¤ S

30. (a) 1 â&#x2030;¤ b

29. 2Ď&#x20AC;k 1 + [f (x)]2 â&#x2030;¤ 2Ď&#x20AC;f (x) 1 + [f (x)]2 â&#x2030;¤ 2Ď&#x20AC;K 1 + [f (x)]2 , so b b b 2Ď&#x20AC;k 1 + [f (x)]2 dx â&#x2030;¤ 2Ď&#x20AC;f (x) 1 + [f (x)]2 dx â&#x2030;¤ 2Ď&#x20AC;K 1 + [f (x)]2 dx,

(b) 2Ď&#x20AC;A = S if f (x) = 0 for all x in [a, b] so f (x) is constant on [a, b].

EXERCISE SET 7.6

1. (a) W = F Âˇ d = 30(7) = 210 ftÂˇlb 6 6 6 1 (b) W = F (x) dx = xâ&#x2C6;&#x2019;2 dx = â&#x2C6;&#x2019; = 5/6 ftÂˇlb x 1 1 1

2. W = 0

40 dx â&#x2C6;&#x2019;

F (x) dx =

3. distance traveled =

40 (x â&#x2C6;&#x2019; 5) dx = 80 + 60 = 140 J 3

v(t) dt =

4t 2 5 dt = t2 = 10 ft. The force is a constant 10 lb, so the 5 5 0

work done is 10 Âˇ 10 = 100 ftÂˇlb.

uh a

4. (a) F (x) = kx, F (0.05) = 0.05k = 45, k = 900 N/m 0.03 900x dx = 0.405 J (c) W = (b) W = 0

5. F (x) = kx, F (0.2) = 0.2k = 100, k = 500 N/m, W =

900x dx = 3.375 J

0.05 0.8

500xdx = 160 J 0

6. F (x) = kx, F (1/2) = k/2 = 6, k = 12 N/m, W =

0.10

12x dx = 24 J 0

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Exercise Set 7.6

303

7. W =

kx dx = k/2 = 10, k = 20 lb/ft 6

(9 â&#x2C6;&#x2019; x)62.4(25Ď&#x20AC;)dx

8. W = 0

(9 â&#x2C6;&#x2019; x)dx = 56, 160Ď&#x20AC; ftÂˇlb

= 1560Ď&#x20AC;

6 9-x

(9 â&#x2C6;&#x2019; x)Ď (25Ď&#x20AC;)dx = 900Ď&#x20AC;Ď ftÂˇlb

9. W = 0

(15x2 â&#x2C6;&#x2019; x3 )dx 0

(3x â&#x2C6;&#x2019; x2 )dx

= 78480

= 261, 600 J

= 3000

3 2 x

4 â&#x2C6;&#x2019; x2 dx â&#x2C6;&#x2019; 1000

â&#x2C6;&#x2019;2

3-x

w(x) 0

3 w(x)

2 3-x x

x 4 â&#x2C6;&#x2019; x2 dx

2 0

â&#x2C6;&#x2019;2

w = 2 4 â&#x2C6;&#x2019; x2 2 (3 â&#x2C6;&#x2019; x)(50)(2 4 â&#x2C6;&#x2019; x2 )(10)dx W = â&#x2C6;&#x2019;2

15 - x

12.

11. w/4 = x/3, w = 4x/3, 2 W = (3 â&#x2C6;&#x2019; x)(9810)(4x/3)(6)dx

10 x

= 208, 000Ď&#x20AC;/3 ftÂˇlb

83.2 Ď&#x20AC; = 3

10. r/10 = x/15, r = 2x/3, 10 W = (15 â&#x2C6;&#x2019; x)62.4(4Ď&#x20AC;x2 /9)dx 0

us uf

= 3000[Ď&#x20AC;(2)2 /2] â&#x2C6;&#x2019; 0 = 6000Ď&#x20AC; ftÂˇlb

13. (a) W =

(10 â&#x2C6;&#x2019; x)62.4(300)dx

uh a

-2

10 9 10 - x x

(10 â&#x2C6;&#x2019; x)dx

= 18,720 0

= 926,640 ftÂˇlb

(b) to empty the pool in one hour would require 926,640/3600 = 257.4 ftÂˇlb of work per second so hp of motor = 257.4/550 = 0.468

0 20

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Chapter 7

14. W =

x(62.4)(300) dx = 18,720 0

x dx = (81/2)18,720 = 758,160 ftÂˇlb 0

100

15(100 â&#x2C6;&#x2019; x)dx

15. W =

us uf

304

Pulley

100

= 75, 000 ftÂˇlb

100 - x

Chain

nâ&#x2020;&#x2019;+â&#x2C6;&#x17E;

16. The total time of winding the rope is (20 ft)/(2 ft/s) = 10 s. During the time interval from time t to time t + â&#x2C6;&#x2020;t the work done is â&#x2C6;&#x2020;W = F (t) Âˇ â&#x2C6;&#x2020;x. The distance â&#x2C6;&#x2020;x = 2â&#x2C6;&#x2020;t, and the force F (t) is given by the weight w(t) of the bucket, rope and water at time t. The bucket and its remaining water together weigh (3 + 20) â&#x2C6;&#x2019; t/2 lb, and the rope is 20 â&#x2C6;&#x2019; 2t ft long and weighs 4(20 â&#x2C6;&#x2019; 2t) oz or 5 â&#x2C6;&#x2019; t/2 lb. Thus at time t the bucket, water and rope together weigh w(t) = 23 â&#x2C6;&#x2019; t/2 + 5 â&#x2C6;&#x2019; t/2 = 28 â&#x2C6;&#x2019; t lb. The amount of work done in the time interval from time t to time t + â&#x2C6;&#x2020;t is thus â&#x2C6;&#x2020;W = (28 â&#x2C6;&#x2019; t)2â&#x2C6;&#x2020;t, and the total work done is 10 10 W = lim (28 â&#x2C6;&#x2019; t)2â&#x2C6;&#x2020;t = (28 â&#x2C6;&#x2019; t)2 dt = 2(28t â&#x2C6;&#x2019; t2 /2) = 460 ftÂˇlb. 0

17. When the rocket is x ft above the ground

3000

total weight = weight of rocket + weight of fuel = 43 â&#x2C6;&#x2019; x/500 tons,

3000

= 3 + [40 â&#x2C6;&#x2019; 2(x/1000)]

x Rocket

(43 â&#x2C6;&#x2019; x/500)dx = 120, 000 ftÂˇtons

W = 0

18. Let F (x) be the force needed to hold charge A at position x, then c c F (x) = , F (â&#x2C6;&#x2019;a) = 2 = k, 2 (a â&#x2C6;&#x2019; x) 4a

A -a

B 0

â&#x2C6;&#x2019;a

so c = 4a2 k. 0 W = 4a2 k(a â&#x2C6;&#x2019; x)â&#x2C6;&#x2019;2 dx = 2ak J

uh a

19. (a) 150 = k/(4000)2 , k = 2.4 Ă&#x2014; 109 , w(x) = k/x2 = 2,400,000,000/x2 lb (b) 6000 = k/(4000)2 , k = 9.6 Ă&#x2014; 1010 , w(x) = 9.6 Ă&#x2014; 1010 /(x + 4000)2 lb 5000 (c) W = 9.6(1010 )xâ&#x2C6;&#x2019;2 dx = 4,800,000 miÂˇlb = 2.5344 Ă&#x2014; 1010 ftÂˇlb 4000

20. (a) 20 = k/(1080)2 , k = 2.3328 Ă&#x2014; 107 , weight = w(x + 1080) = 2.3328 Âˇ 107 /(x + 1080)2 lb 10.8 (b) W = [2.3328 Âˇ 107 /(x + 1080)2 ] dx = 213.86 miÂˇlb = 1,129,188 ftÂˇlb 0

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Exercise Set 7.7

305

21. W = F Âˇ d = (6.40 Ă&#x2014; 105 )(3.00 Ă&#x2014; 103 ) = 1.92 Ă&#x2014; 109 J; from the Work-Energy Relationship (5),

us uf

vf2 = 2W/m + vi2 = 2(1.92 Âˇ 109 )/(4 Âˇ 105 ) + 202 = 10,000, vf = 100 m/s 22. W = F Âˇ d = (2.00 Ă&#x2014; 105 )(2.00 Ă&#x2014; 105 ) = 4 Ă&#x2014; 1010 J; from the Work-Energy Relationship (5), vf2 = 2W/m + vi2 = 8 Âˇ 1010 /(2 Âˇ 103 ) + 108 â&#x2030;&#x2C6; 11.832 m/s. 23. (a) The kinetic energy would have decreased by

(b) (4.5 Ă&#x2014; 1014 )/(4.2 Ă&#x2014; 1015 ) â&#x2030;&#x2C6; 0.107

1 1 mv 2 = 4 Âˇ 106 (15000)2 = 4.5 Ă&#x2014; 1014 J 2 2 1000 (c) (0.107) â&#x2030;&#x2C6; 8.24 bombs 13

EXERCISE SET 7.7

(b) F = Ď hA = 9810(10)(25) = 2,452,500 N P = Ď h = 9810(10) = 98.1 kPa

1. (a) F = Ď hA = 62.4(5)(100) = 31,200 lb 2

P = Ď h = 62.4(5) = 312 lb/ft

2. (a) F = P A = 6 Âˇ 105 (160) = 9.6 Ă&#x2014; 107 N

(b) F = P A = 100(60) = 6000 lb

62.4x(4)dx 0

4. F =

= 249.6

x dx = 499.2 lb 0

9810x(2 25 â&#x2C6;&#x2019; x2 )dx

x(25 â&#x2C6;&#x2019; x2 )1/2 dx

= 19,620

= 8.175 Ă&#x2014; 105 N

x dx 1

y = â&#x2C6;&#x161;25 â&#x20AC;&#x201C; x 2

2â&#x2C6;&#x161;25 â&#x20AC;&#x201C; x 2

0 4

1 x 3

6. By similar triangles â&#x2C6;&#x161; 2 â&#x2C6;&#x161; w(x) 2 3â&#x2C6;&#x2019;x â&#x2C6;&#x161; , w(x) = â&#x2C6;&#x161; (2 3 â&#x2C6;&#x2019; x), = 4 2 3 3

2â&#x2C6;&#x161;3 2 â&#x2C6;&#x161; 62.4x â&#x2C6;&#x161; (2 3 â&#x2C6;&#x2019; x) dx F = 3 0 2â&#x2C6;&#x161;3 â&#x2C6;&#x161; 124.8 (2 3x â&#x2C6;&#x2019; x2 )dx = 499.2 lb = â&#x2C6;&#x161; 3 0 4

w(x)

x 4

2 â&#x2C6;&#x161;3

uh a

= 39,240

= 156,960 N

5. F =

9810x(4)dx

3. F =

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Chapter 7

7. By similar triangles w(x) 10 â&#x2C6;&#x2019; x = 6 8 3 w(x) = (10 â&#x2C6;&#x2019; x), 4

10 3 F = 9810x (10 â&#x2C6;&#x2019; x) dx 4 2 10 (10x â&#x2C6;&#x2019; x2 )dx = 1,098,720 N = 7357.5

8. w(x) = 16 + 2u(x), but

us uf

w(x) = 16 + (12 â&#x2C6;&#x2019; x) = 28 â&#x2C6;&#x2019; x, 12 F = 62.4x(28 â&#x2C6;&#x2019; x)dx 12

(28x â&#x2C6;&#x2019; x2 )dx = 77,209.6 lb.

= 62.4

0 6

4 x

u(x)

4 w(x)

1 12 â&#x2C6;&#x2019; x u(x) so u(x) = (12 â&#x2C6;&#x2019; x), = 8 2 4

w(x)

306

9. Yes: if Ď 2 = 2Ď 1 then F2 =

Ď 2 h(x)w(x) dx = a

50x(2 4 â&#x2C6;&#x2019; x2 )dx

x(4 â&#x2C6;&#x2019; x2 )1/2 dx

= 100 0

Ď 1 h(x)w(x) dx = 2F1 . a

2 y

y = â&#x2C6;&#x161;4 â&#x20AC;&#x201C; x2 x

2â&#x2C6;&#x161;4 â&#x20AC;&#x201C; x2

= 800/3 lb

2Ď 1 h(x)w(x) dx = 2

10. F =

11. Find the forces on the upper and lower halves and add them:

â&#x2C6;&#x161; 2a/2

w1 (x) x â&#x2C6;&#x161; =â&#x2C6;&#x161; , w1 (x) = 2x 2a 2a/2 â&#x2C6;&#x161;2a/2 Ď x(2x)dx = 2Ď F1 = 0

x dx =

â&#x2C6;&#x161;

0 3

2Ď a /6,

â&#x2C6;&#x161; â&#x2C6;&#x161; w2 (x) 2a â&#x2C6;&#x2019; x â&#x2C6;&#x161; = â&#x2C6;&#x161; , w2 (x) = 2( 2a â&#x2C6;&#x2019; x) 2a 2a/2 â&#x2C6;&#x161;2a â&#x2C6;&#x161;2a â&#x2C6;&#x161; â&#x2C6;&#x161; â&#x2C6;&#x161; Ď x[2( 2a â&#x2C6;&#x2019; x)]dx = 2Ď â&#x2C6;&#x161; ( 2ax â&#x2C6;&#x2019; x2 )dx = 2Ď a3 /3, F2 = â&#x2C6;&#x161;

2a/2

F = F1 + F2 =

â&#x2C6;&#x161;

2Ď a3 /6 +

â&#x2C6;&#x161;

x â&#x2C6;&#x161;2a/2 x

w1(x) a

a â&#x2C6;&#x161;2a w2(x)

â&#x2C6;&#x161;2a

2a/2

â&#x2C6;&#x161; 2Ď a3 /3 = Ď a3 / 2 lb

uh a

12. If a constant vertical force is applied to a ďŹ&#x201A;at plate which is horizontal and the magnitude of the force is F , then, if the plate is tilted so as to form an angle Î¸ with the vertical, the magnitude of the force on the plate decreases to F cos Î¸.

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Exercise Set 7.7

307

us uf

Suppose that a flat surface is immersed, at an angle θ with the vertical, in a fluid of weight density ρ, and that the submerged portion of the surface extends from x = a to x = b along an x-axis whose positive diretion is not necessarily down, but is slanted. Following the derivation of equation (8), we divide the interval [a, b] into n subintervals a = x0 < x1 < . . . < xn−1 < xn = b. Then the magnitude Fk of the force on the plate satisfies the inequalities ρh(xk−1 )Ak cos θ ≤ Fk ≤ ρh(xk )Ak cos θ, or equivalently that Fk sec θ ≤ h(xk ). Following the argument in the text we arrive at the desired equation h(xk−1 ) ≤ ρAk b F = ρh(x)w(x) sec θ dx. √

√ √ 162 + 42 = 272 = 4 17 is the other dimension of the bottom. √ (h(x) − 4)/4 = x/(4 17) √ h(x) = x/ 17 + 4, √ √ sec θ = 4 17/16 = 17/4 F =

4 0 h(x)

√ √ 62.4(x/ 17 + 4)10( 17/4) dx

√ = 156 17

√ 4 17

13.

4√17

√ (x/ 17 + 4)dx

= 63,648 lb

14. If we lower the water level √ by y ft then the force F1 is computed as in Exercise 13, but with h(x) replaced by h1 (x) = x/ 17 + 4 − y, and we obtain 4√17 √ 62.4(10) 17/4 dx = F − 624(17)y = 63,648 − 10,608y. F1 = F − y 0

If F1 = F/2 then 63,648/2 = 63,648 − 10,608y, y = 63,648/(2 · 10,608) = 3, so the water level should be reduced by 3 ft. 15. h(x) = x sin 60◦ =

√ √

3x/2,

200

θ = 30◦ , sec θ = 2/ 3, 100 √ √ F = 9810( 3x/2)(200)(2/ 3) dx

h(x)

60° 100

100

= 200 · 9810

x dx

= 9810 · 1003 = 9.81 × 109 N

h+2

16. F =

ρ0 x(2)dx

uh a

h+2

= 2ρ0

x dx

= 4ρ0 (h + 1)

h x h+2

2 2

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308

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Chapter 7

us uf

17. (a) From Exercise 16, F = 4ρ0 (h + 1) so (assuming that ρ0 is constant) dF/dt = 4ρ0 (dh/dt) which is a positive constant if dh/dt is a positive constant. (b) If dh/dt = 20 then dF/dt = 80ρ0 lb/min from Part (a). 18. (a) Let h1 and h2 be the maximum and minimum depths of the disk Dr . The pressure P (r) on one side of the disk satisﬁes inequality (5): ρh1 ≤ P (r) ≤ ρh2 . But lim h1 = lim h2 = h, and hence

r→0+

ρh = lim ρh1 ≤ lim P (r) ≤ lim ρh2 = ρh, so lim P (r) = ρh. r→0+

r→0+

(b) The disks Dr in Part (a) have no particular direction (the axes of the disks have arbitrary direction). Thus P , the limiting value of P (r), is independent of direction.

EXERCISE SET 7.8 1. (a) sinh 3 ≈ 10.0179 (b) cosh(−2) ≈ 3.7622

2. (a) csch(−1) ≈ −0.8509 (b) sech(ln 2) = 0.8

1 4 3− = 3 3 5 1 1 1 +2 = (b) cosh(− ln 2) = (e− ln 2 + eln 2 ) = 2 2 2 4

1 1 3. (a) sinh(ln 3) = (eln 3 − e− ln 3 ) = 2 2

e2 ln 5 − e−2 ln 5 25 − 1/25 312 = = 2 ln 5 −2 ln 5 e +e 25 + 1/25 313 63 1 1 1 −3 ln 2 3 ln 2 −8 =− −e )= (d) sinh(−3 ln 2) = (e 2 2 8 16 (c) tanh(2 ln 5) =

1 x+ x

1 ln x 1 (e + e− ln x ) = 2 2

4. (a)

1 1 ln x − e− ln x ) = (e (b) 2 2

1 x− x

x2 + 1 ,x>0 2x

x2 − 1 ,x>0 2x

uh a

x2 − 1/x2 x4 − 1 e2 ln x − e−2 ln x = = ,x>0 e2 ln x + e−2 ln x x2 + 1/x2 x4 + 1 1 + x2 1 1 1 − ln x ln x (e +x = ,x>0 +e )= (d) 2 2 x 2x

(c)

sinh x0

cosh x0

tanh x0

coth x0

sech x0

csch x0

(a)

√5

2 /√5

√5 / 2

1/√5

1/ 2

(b)

3/ 4

5/ 4

3/ 5

5/3

4/5

4/3

(c)

4/3

5/ 3

4/5

5/4

3/ 5

3/ 4

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Exercise Set 7.8

(a) cosh2 x0 = 1 + sinh2 x0 = 1 + (2)2 = 5, cosh x0 =

â&#x2C6;&#x161;

309

9 25 3 â&#x2C6;&#x2019;1= , sinh x0 = (because x0 > 0) 16 16 4 2 9 4 16 3 = , sech x0 = , =1â&#x2C6;&#x2019; (c) sech2 x0 = 1 â&#x2C6;&#x2019; tanh2 x0 = 1 â&#x2C6;&#x2019; 5 25 25 5 4 4 5 1 5 sinh x0 cosh x0 = = = tanh x0 we get sinh x0 = = , from 3 5 3 sech x0 3 cosh x0

us uf

d 1 cosh x d cschx = =â&#x2C6;&#x2019; = â&#x2C6;&#x2019; coth x csch x for x = 0 dx dx sinh x sinh2 x sinh x 1 d d =â&#x2C6;&#x2019; = â&#x2C6;&#x2019; tanh x sech x for all x sech x = dx dx cosh x cosh2 x

sinh2 x â&#x2C6;&#x2019; cosh2 x d cosh x d = coth x = = â&#x2C6;&#x2019; csch2 x for x = 0 dx sinh x dx sinh2 x dy dy dx = cosh y; so dx dy dx d dy 1 1 1 for all x. =â&#x2C6;&#x161; [sinhâ&#x2C6;&#x2019;1 x] = = = 2 dx dx cosh y 1 + x2 1 + sinh y

7. (a) y = sinhâ&#x2C6;&#x2019;1 x if and only if x = sinh y; 1 =

(b) Let x â&#x2030;Ľ 1. Then y = coshâ&#x2C6;&#x2019;1 x if and only if x = cosh y; 1 =

dy dx dy = sinh y, so dx dy dx

1 1 d dy 1 [coshâ&#x2C6;&#x2019;1 x] = = = for x â&#x2030;Ľ 1. = 2 2 dx dx sinh y x â&#x2C6;&#x2019;1 cosh y â&#x2C6;&#x2019; 1

(c) Let â&#x2C6;&#x2019;1 < x < 1. Then y = tanhâ&#x2C6;&#x2019;1 x if and only if x = tanh y; thus dy dx dy dy d dy 1 1= = . sech2 y = (1 â&#x2C6;&#x2019; tanh2 y) = 1 â&#x2C6;&#x2019; x2 , so [tanhâ&#x2C6;&#x2019;1 x] = = dx dy dx dx dx dx 1 â&#x2C6;&#x2019; x2 9. 4 cosh(4x â&#x2C6;&#x2019; 8) sech2 2x tanh 2x

13.

1 csch(1/x) coth(1/x) x2

2 + 5 cosh(5x) sinh(5x) 4x + cosh2 (5x)

14. â&#x2C6;&#x2019;2e2x sech(e2x ) tanh(e2x )

16. 6 sinh2 (2x) cosh(2x)

15.

12. 2

1 11. â&#x2C6;&#x2019; csch2 (ln x) x

10. 4x3 sinh(x4 )

â&#x2C6;&#x161; â&#x2C6;&#x161; â&#x2C6;&#x161; 17. x5/2 tanh( x) sech2 ( x) + 3x2 tanh2 ( x)

1/x2

uh a

20.

18. â&#x2C6;&#x2019;3 cosh(cos 3x) sin 3x

(â&#x2C6;&#x2019;1/x2 ) = â&#x2C6;&#x2019;

19. 1 â&#x2C6;&#x161; |x| x2 + 1

â&#x2C6;&#x161; â&#x2C6;&#x2019;1 2 2 22. 1/ (sinh x) â&#x2C6;&#x2019; 1 1 + x

â&#x2C6;&#x2019;1

24. 2(coth

x)/(1 â&#x2C6;&#x2019; x ) 2

1 = 1/ 9 + x2 2 1 + x /9 3 1

â&#x2C6;&#x161;

21. 1/ (coshâ&#x2C6;&#x2019;1 x) x2 â&#x2C6;&#x2019; 1

23. â&#x2C6;&#x2019;(tanhâ&#x2C6;&#x2019;1 x)â&#x2C6;&#x2019;2 /(1 â&#x2C6;&#x2019; x2 )

25.

sinh x

(b) sinh2 x0 = cosh2 x0 â&#x2C6;&#x2019; 1 =

sinh x = = 2 | sinh x| cosh x â&#x2C6;&#x2019; 1

1, x > 0 â&#x2C6;&#x2019;1, x < 0

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Chapter 7

1 + tanh2 x

x â&#x2C6;&#x2019;1 â&#x2C6;&#x161; â&#x2C6;&#x2019; + csch x |x| 1 + x2

1 sinh7 x + C 7

32.

1 34. â&#x2C6;&#x2019; coth(3x) + C 3

43.

44.

1 1 du = sinhâ&#x2C6;&#x2019;1 3x + C 3 1 + u2 â&#x2C6;&#x161; â&#x2C6;&#x161; â&#x2C6;&#x161; 2 1 â&#x2C6;&#x161; â&#x2C6;&#x161; x = 2u, du = du = coshâ&#x2C6;&#x2019;1 (x/ 2) + C 2u2 â&#x2C6;&#x2019; 2 u2 â&#x2C6;&#x2019; 1 1 â&#x2C6;&#x161; du = â&#x2C6;&#x2019; sechâ&#x2C6;&#x2019;1 (ex ) + C u = ex , u 1 â&#x2C6;&#x2019; u2 1 u = cos Î¸, â&#x2C6;&#x2019; â&#x2C6;&#x161; du = â&#x2C6;&#x2019; sinhâ&#x2C6;&#x2019;1 (cos Î¸) + C 1 + u2 du â&#x2C6;&#x161; = â&#x2C6;&#x2019;cschâ&#x2C6;&#x2019;1 |u| + C = â&#x2C6;&#x2019;cschâ&#x2C6;&#x2019;1 |2x| + C u = 2x, u 1 + u2 1 5/3 1 1 â&#x2C6;&#x161; â&#x2C6;&#x161; x = 5u/3, du = du = coshâ&#x2C6;&#x2019;1 (3x/5) + C 2 2 3 3 25u â&#x2C6;&#x2019; 25 u â&#x2C6;&#x2019;1 1 3

â&#x2C6;&#x161;

42.

ln 3 = ln 5 â&#x2C6;&#x2019; ln 3 38. ln(cosh x)

41.

2 (tanh x)3/2 + C 3

1 36. â&#x2C6;&#x2019; coth3 x + C 3

35. ln(cosh x) + C

ln 3 1 3 â&#x2C6;&#x2019; sech x = 37/375 3 ln 2

39. u = 3x,

40.

33.

37.

1 sinh(2x â&#x2C6;&#x2019; 3) + C 2

31.

ex â&#x2C6;&#x161; + ex sechâ&#x2C6;&#x2019;1 x 2x 1 â&#x2C6;&#x2019; x

28. 10(1 + x csch

â&#x2C6;&#x2019;1

27. â&#x2C6;&#x2019;

us uf

26. (sech2 x)/

310

1/2 1 1 + 1/2 1 45. tanhâ&#x2C6;&#x2019;1 x = tanhâ&#x2C6;&#x2019;1 (1/2) â&#x2C6;&#x2019; tanhâ&#x2C6;&#x2019;1 (0) = ln = ln 3 2 1 â&#x2C6;&#x2019; 1/2 2 0 46. sinh

â&#x2C6;&#x161;3 0

ln 3

49. A =

â&#x2C6;&#x161;

â&#x2C6;&#x161; 3 â&#x2C6;&#x2019; sinhâ&#x2C6;&#x2019;1 0 = ln( 3 + 2)

ln 3 1 1 sinh 2x dx = cosh 2x = [cosh(2 ln 3) â&#x2C6;&#x2019; 1], 2 2 0

= sinhâ&#x2C6;&#x2019;1

â&#x2C6;&#x2019;1

1 1 1 ln 9 (e + eâ&#x2C6;&#x2019; ln 9 ) = (9 + 1/9) = 41/9 so A = [41/9 â&#x2C6;&#x2019; 1] = 16/9. 2 2 2 ln 2 ln 2 50. V = Ď&#x20AC; sech2 x dx = Ď&#x20AC; tanh x = Ď&#x20AC; tanh(ln 2) = 3Ď&#x20AC;/5

but cosh(2 ln 3) = cosh(ln 9) =

uh a

51. V = Ď&#x20AC;

52.

(cosh2 2x â&#x2C6;&#x2019; sinh2 2x)dx = Ď&#x20AC;

dx = 5Ď&#x20AC; 0

1 1 1 cosh ax dx = 2, sinh ax = 2, sinh a = 2, sinh a = 2a; a a 0

let f (a) = sinh a â&#x2C6;&#x2019; 2a, then an+1 = an â&#x2C6;&#x2019;

sinh an â&#x2C6;&#x2019; 2an , a1 = 2.2, . . . , a4 = a5 = 2.177318985. cosh an â&#x2C6;&#x2019; 2

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us uf

53. y = sinh x, 1 + (y )2 = 1 + sinh2 x = cosh2 x ln 2 ln 2 1 3 1 ln 2 1 â&#x2C6;&#x2019; ln 2 2â&#x2C6;&#x2019; = L= cosh x dx = sinh x = sinh(ln 2) = (e â&#x2C6;&#x2019; e )= 2 2 2 4 0 0 54. y = sinh(x/a), 1 + (y )2 = 1 + sinh2 (x/a) = cosh2 (x/a) x1 x1 cosh(x/a)dx = a sinh(x/a) = a sinh(x1 /a) L= 0

cosh(â&#x2C6;&#x2019;x) =

1 â&#x2C6;&#x2019;x 1 (e â&#x2C6;&#x2019; ex ) = â&#x2C6;&#x2019; (ex â&#x2C6;&#x2019; eâ&#x2C6;&#x2019;x ) = â&#x2C6;&#x2019; sinh x 2 2

55. sinh(â&#x2C6;&#x2019;x) =

311

Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Exercise Set 7.8

1 â&#x2C6;&#x2019;x 1 (e + ex ) = (ex + eâ&#x2C6;&#x2019;x ) = cosh x 2 2

1 1 x (e + eâ&#x2C6;&#x2019;x ) + (ex â&#x2C6;&#x2019; eâ&#x2C6;&#x2019;x ) = ex 2 2 1 1 x (b) cosh x â&#x2C6;&#x2019; sinh x = (e + eâ&#x2C6;&#x2019;x ) â&#x2C6;&#x2019; (ex â&#x2C6;&#x2019; eâ&#x2C6;&#x2019;x ) = eâ&#x2C6;&#x2019;x 2 2 1 x 1 (c) sinh x cosh y + cosh x sinh y = (e â&#x2C6;&#x2019; eâ&#x2C6;&#x2019;x )(ey + eâ&#x2C6;&#x2019;y ) + (ex + eâ&#x2C6;&#x2019;x )(ey â&#x2C6;&#x2019; eâ&#x2C6;&#x2019;y ) 4 4 =

56. (a) cosh x + sinh x =

1 x+y [(e â&#x2C6;&#x2019; eâ&#x2C6;&#x2019;x+y + exâ&#x2C6;&#x2019;y â&#x2C6;&#x2019; eâ&#x2C6;&#x2019;xâ&#x2C6;&#x2019;y ) + (ex+y + eâ&#x2C6;&#x2019;x+y â&#x2C6;&#x2019; exâ&#x2C6;&#x2019;y â&#x2C6;&#x2019; eâ&#x2C6;&#x2019;xâ&#x2C6;&#x2019;y )] 4

1 (x+y) â&#x2C6;&#x2019; eâ&#x2C6;&#x2019;(x+y) ] = sinh(x + y) [e 2 (d) Let y = x in Part (c). =

Let y = x in Part (e).

(f )

(e) The proof is similar to Part (c), or: treat x as variable and y as constant, and diďŹ&#x20AC;erentiate the result in Part (c) with respect to x. (g) Use cosh2 x = 1 + sinh2 x together with Part (f). (h) Use sinh2 x = cosh2 x â&#x2C6;&#x2019; 1 together with Part (f). 57. (a) Divide cosh2 x â&#x2C6;&#x2019; sinh2 x = 1 by cosh2 x.

sinh x sinh y + sinh x cosh y + cosh x sinh y tanh x + tanh y cosh x cosh y (b) tanh(x + y) = = = sinh x sinh y cosh x cosh y + sinh x sinh y 1 + tanh x tanh y 1+ cosh x cosh y (c) Let y = x in Part (b).

uh a

1 58. (a) Let y = coshâ&#x2C6;&#x2019;1 x; then x = cosh y = (ey + eâ&#x2C6;&#x2019;y ), ey â&#x2C6;&#x2019; 2x + eâ&#x2C6;&#x2019;y = 0, e2y â&#x2C6;&#x2019; 2xey + 1 = 0, 2 â&#x2C6;&#x161; 2x Âą 4x2 â&#x2C6;&#x2019; 4 y 2 e = = x Âą x â&#x2C6;&#x2019; 1. To determine which sign to take, note that y â&#x2030;Ľ 0 2 â&#x2C6;&#x161; so eâ&#x2C6;&#x2019;y â&#x2030;¤ ey , x = (ey + eâ&#x2C6;&#x2019;y )/2 â&#x2030;¤ (ey + ey )/2 = ey , hence ey â&#x2030;Ľ x thus ey = x + x2 â&#x2C6;&#x2019; 1, â&#x2C6;&#x161; y = coshâ&#x2C6;&#x2019;1 x = ln(x + x2 â&#x2C6;&#x2019; 1). ey â&#x2C6;&#x2019; eâ&#x2C6;&#x2019;y e2y â&#x2C6;&#x2019; 1 = , xe2y + x = e2y â&#x2C6;&#x2019; 1, ey + eâ&#x2C6;&#x2019;y e2y + 1 1+x 1 1+x = (1 + x)/(1 â&#x2C6;&#x2019; x), 2y = ln , y = ln . 1â&#x2C6;&#x2019;x 2 1â&#x2C6;&#x2019;x

(b) Let y = tanhâ&#x2C6;&#x2019;1 x; then x = tanh y = 1 + x = e2y (1 â&#x2C6;&#x2019; x), e2y

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312

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Chapter 7

â&#x2C6;&#x161; d 1 + x/ x2 â&#x2C6;&#x2019; 1 â&#x2C6;&#x161; (coshâ&#x2C6;&#x2019;1 x) = = 1/ x2 â&#x2C6;&#x2019; 1 dx x + x2 â&#x2C6;&#x2019; 1

1 d d 1 1 1 = 1/(1 â&#x2C6;&#x2019; x2 ) (tanhâ&#x2C6;&#x2019;1 x) = (b) (ln(1 + x) â&#x2C6;&#x2019; ln(1 â&#x2C6;&#x2019; x)) = + dx 2 2 1+x 1â&#x2C6;&#x2019;x dx

us uf

59. (a)

60. Let y = sechâ&#x2C6;&#x2019;1 x then x = sech y = 1/ cosh y, cosh y = 1/x, y = coshâ&#x2C6;&#x2019;1 (1/x); the proofs for the remaining two are similar.

For |u| > 1,

du = cothâ&#x2C6;&#x2019;1 u + C = tanhâ&#x2C6;&#x2019;1 (1/u) + C. 1 â&#x2C6;&#x2019; u2

â&#x2C6;&#x161; d x d 1 1 â&#x2C6;&#x161; =â&#x2C6;&#x2019; â&#x2C6;&#x161; (sechâ&#x2C6;&#x2019;1 |x|) = (sechâ&#x2C6;&#x2019;1 x2 ) = â&#x2C6;&#x2019; â&#x2C6;&#x161; â&#x2C6;&#x161; 2 2 2 dx dx x 1 â&#x2C6;&#x2019; x2 x 1â&#x2C6;&#x2019;x x

(b) Similar to solution of Part (a)

(d) (e) (f )

(c)

xâ&#x2020;&#x2019;+â&#x2C6;&#x17E;

lim tanhâ&#x2C6;&#x2019;1 x = lim

xâ&#x2020;&#x2019;1â&#x2C6;&#x2019;

(b)

1 x (e â&#x2C6;&#x2019; eâ&#x2C6;&#x2019;x ) = +â&#x2C6;&#x17E; â&#x2C6;&#x2019; 0 = +â&#x2C6;&#x17E; 2 1 x lim sinh x = lim (e â&#x2C6;&#x2019; eâ&#x2C6;&#x2019;x ) = 0 â&#x2C6;&#x2019; â&#x2C6;&#x17E; = â&#x2C6;&#x2019;â&#x2C6;&#x17E; xâ&#x2020;&#x2019;â&#x2C6;&#x2019;â&#x2C6;&#x17E; xâ&#x2020;&#x2019;â&#x2C6;&#x2019;â&#x2C6;&#x17E; 2 ex â&#x2C6;&#x2019; eâ&#x2C6;&#x2019;x =1 lim tanh x = lim x xâ&#x2020;&#x2019;+â&#x2C6;&#x17E; xâ&#x2020;&#x2019;+â&#x2C6;&#x17E; e + eâ&#x2C6;&#x2019;x ex â&#x2C6;&#x2019; eâ&#x2C6;&#x2019;x = â&#x2C6;&#x2019;1 lim tanh x = lim x xâ&#x2020;&#x2019;â&#x2C6;&#x2019;â&#x2C6;&#x17E; xâ&#x2020;&#x2019;â&#x2C6;&#x2019;â&#x2C6;&#x17E; e + eâ&#x2C6;&#x2019;x lim sinhâ&#x2C6;&#x2019;1 x = lim ln(x + x2 + 1) = +â&#x2C6;&#x17E; lim sinh x = lim

xâ&#x2020;&#x2019;+â&#x2C6;&#x17E;

1 [ln(1 + x) â&#x2C6;&#x2019; ln(1 â&#x2C6;&#x2019; x)] = +â&#x2C6;&#x17E; 2

63. (a)

62. (a)

du = tanhâ&#x2C6;&#x2019;1 u + C. 1 â&#x2C6;&#x2019; u2

61. If |u| < 1 then, by Theorem 7.8.6,

x2 â&#x2C6;&#x2019; 1) â&#x2C6;&#x2019; ln x] xâ&#x2020;&#x2019;+â&#x2C6;&#x17E; xâ&#x2020;&#x2019;+â&#x2C6;&#x17E; â&#x2C6;&#x161; x + x2 â&#x2C6;&#x2019; 1 = lim ln = lim ln(1 + 1 â&#x2C6;&#x2019; 1/x2 ) = ln 2 xâ&#x2020;&#x2019;+â&#x2C6;&#x17E; xâ&#x2020;&#x2019;+â&#x2C6;&#x17E; x cosh x ex + eâ&#x2C6;&#x2019;x 1 = lim = lim (b) lim (1 + eâ&#x2C6;&#x2019;2x ) = 1/2 x x xâ&#x2020;&#x2019;+â&#x2C6;&#x17E; xâ&#x2020;&#x2019;+â&#x2C6;&#x17E; xâ&#x2020;&#x2019;+â&#x2C6;&#x17E; e 2e 2 lim (coshâ&#x2C6;&#x2019;1 x â&#x2C6;&#x2019; ln x) = lim [ln(x +

64. (a)

65. For |x| < 1, y = tanhâ&#x2C6;&#x2019;1 x is deďŹ ned and dy/dx = 1/(1 â&#x2C6;&#x2019; x2 ) > 0; y = 2x/(1 â&#x2C6;&#x2019; x2 )2 changes sign at x = 0, so there is a point of inďŹ&#x201A;ection there.

uh a

1 a â&#x2C6;&#x161; 66. Let x = â&#x2C6;&#x2019;u/a, du = â&#x2C6;&#x2019; dx = â&#x2C6;&#x2019; coshâ&#x2C6;&#x2019;1 x + C = â&#x2C6;&#x2019; coshâ&#x2C6;&#x2019;1 (â&#x2C6;&#x2019;u/a) + C. 2 2 2 u â&#x2C6;&#x2019;a a x â&#x2C6;&#x2019;1 â&#x2C6;&#x161; 2 â&#x2C6;&#x2019; a2 u + u a â&#x2C6;&#x2019;1 â&#x2C6;&#x161; â&#x2C6;&#x161; â&#x2C6;&#x2019; cosh (â&#x2C6;&#x2019;u/a) = â&#x2C6;&#x2019; ln(â&#x2C6;&#x2019;u/a + u2 /a2 â&#x2C6;&#x2019; 1) = ln â&#x2C6;&#x2019;u + u2 â&#x2C6;&#x2019; a2 u + u2 â&#x2C6;&#x2019; a2 = ln u + u2 â&#x2C6;&#x2019; a2 â&#x2C6;&#x2019; ln a = ln |u + u2 â&#x2C6;&#x2019; a2 | + C1 1 â&#x2C6;&#x161; so du = ln u + u2 â&#x2C6;&#x2019; a2 + C2 . u2 â&#x2C6;&#x2019; a2 â&#x2C6;&#x161;

67. Using sinh x + cosh x = ex (Exercise 56a), (sinh x + cosh x)n = (ex )n = enx = sinh nx + cosh nx.

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Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Chapter 7 Supplementary Exercises a

1 tx e t

etx dx = â&#x2C6;&#x2019;a

a = â&#x2C6;&#x2019;a

1 at 2 sinh at (e â&#x2C6;&#x2019; eâ&#x2C6;&#x2019;at ) = for t = 0. t t

68.

313

us uf

69. (a) y = sinh(x/a), 1 + (y )2 = 1 + sinh2 (x/a) = cosh2 (x/a) b b L=2 cosh(x/a) dx = 2a sinh(x/a) = 2a sinh(b/a) 0

(b) The highest point is at x = b, the lowest at x = 0, so S = a cosh(b/a) â&#x2C6;&#x2019; a cosh(0) = a cosh(b/a) â&#x2C6;&#x2019; a.

70. From Part (a) of Exercise 69, L = 2a sinh(b/a) so 120 = 2a sinh(50/a), a sinh(50/a) = 60. Let u = 50/a, then a = 50/u so (50/u) sinh u = 60, sinh u = 1.2u. If f (u) = sinh u â&#x2C6;&#x2019; 1.2u, then sinh un â&#x2C6;&#x2019; 1.2un ; u1 = 1, . . . , u5 = u6 = 1.064868548 â&#x2030;&#x2C6; 50/a so a â&#x2030;&#x2C6; 46.95415231. un+1 = un â&#x2C6;&#x2019; cosh un â&#x2C6;&#x2019; 1.2 From Part (b), S = a cosh(b/a) â&#x2C6;&#x2019; a â&#x2030;&#x2C6; 46.95415231[cosh(1.064868548) â&#x2C6;&#x2019; 1] â&#x2030;&#x2C6; 29.2 ft.

71. From Part (b) of Exercise 69, S = a cosh(b/a) â&#x2C6;&#x2019; a so 30 = a cosh(200/a) â&#x2C6;&#x2019; a. Let u = 200/a, then a = 200/u so 30 = (200/u)[cosh u â&#x2C6;&#x2019; 1], cosh u â&#x2C6;&#x2019; 1 = 0.15u. If f (u) = cosh u â&#x2C6;&#x2019; 0.15u â&#x2C6;&#x2019; 1, cosh un â&#x2C6;&#x2019; 0.15un â&#x2C6;&#x2019; 1 then un+1 = un â&#x2C6;&#x2019; ; u1 = 0.3, . . . , u4 = u5 = 0.297792782 â&#x2030;&#x2C6; 200/a so sinh un â&#x2C6;&#x2019; 0.15 a â&#x2030;&#x2C6; 671.6079505. From Part (a), L = 2a sinh(b/a) â&#x2030;&#x2C6; 2(671.6079505) sinh(0.297792782) â&#x2030;&#x2C6; 405.9 ft.

72. (a) When the bow of the boat is at the point (x, y) and the person has walked a distance D, then the person is located at the point (0, D), the line segment connecting (0, D) and (x, y) â&#x2C6;&#x161; has length a; thus a2 = x2 + (D â&#x2C6;&#x2019; y)2 , D = y + a2 â&#x2C6;&#x2019; x2 = a sechâ&#x2C6;&#x2019;1 (x/a). 1 + 5/9 â&#x2C6;&#x2019;1 â&#x2030;&#x2C6; 14.44 m. (b) Find D when a = 15, x = 10: D = 15 sech (10/15) = 15 ln 2/3

2 1 2 a a2 x 1 â&#x2C6;&#x2019; +x =â&#x2C6;&#x2019; (c) dy/dx = â&#x2C6;&#x2019; â&#x2C6;&#x161; +â&#x2C6;&#x161; =â&#x2C6;&#x161; a â&#x2C6;&#x2019; x2 , 2 2 2 2 2 2 x x a â&#x2C6;&#x2019;x a â&#x2C6;&#x2019;x x a â&#x2C6;&#x2019;x 15 15 225 a2 â&#x2C6;&#x2019; x2 a2 225 2 1 + [y ] = 1 + = 2 ; with a = 15 and x = 5, L = dx = â&#x2C6;&#x2019; = 30 m. x2 x x2 x 5 5

CHAPTER 7 SUPPLEMENTARY EXERCISES 2

(2 + x â&#x2C6;&#x2019; x2 ) dx

(b) A =

6. (a) A =

â&#x2C6;&#x161;

y dy +

â&#x2C6;&#x161; [( y â&#x2C6;&#x2019; (y â&#x2C6;&#x2019; 2)] dy

[(2 + x)2 â&#x2C6;&#x2019; x4 ] dx 2 4 â&#x2C6;&#x161; â&#x2C6;&#x161; (d) V = 2Ď&#x20AC; y y dy + 2Ď&#x20AC; y[ y â&#x2C6;&#x2019; (y â&#x2C6;&#x2019; 2)] dy 0 2 2 (e) V = 2Ď&#x20AC; x(2 + x â&#x2C6;&#x2019; x2 ) dx (f ) V = Ď&#x20AC; (c) V = Ď&#x20AC;

uh a

(f (x) â&#x2C6;&#x2019; g(x)) dx + a

(b) A = â&#x2C6;&#x2019;1

(x3 â&#x2C6;&#x2019; x) dx +

(g(x) â&#x2C6;&#x2019; f (x)) dx + b

(x â&#x2C6;&#x2019; x3 ) dx + 0

(f (x) â&#x2C6;&#x2019; g(x)) dx c

(x3 â&#x2C6;&#x2019; x) dx = 1

Ď&#x20AC;(y â&#x2C6;&#x2019; (y â&#x2C6;&#x2019; 2)2 ) dy

y dy +

7. (a) A =

1 1 9 11 + + = 4 4 4 4

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Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Chapter 7

8/27

8. (a) S = 0

2Ď&#x20AC;x 1 + xâ&#x2C6;&#x2019;4/3 dx

(b) S =

2Ď&#x20AC; 0

2Ď&#x20AC;(y + 2) 1 + y 4 /81 dy

y3 1 + y 4 /81 dy 27

314

us uf

2 y 1/3 y 2/3 dy dy x2/3 + y 2/3 a2/3 =â&#x2C6;&#x2019; 9. By implicit diďŹ&#x20AC;erentiation , so 1 + =1+ = = , dx x dx x x2/3 x2/3 â&#x2C6;&#x2019;a/8 â&#x2C6;&#x2019;a/8 a1/3 1/3 L= dx = â&#x2C6;&#x2019;a xâ&#x2C6;&#x2019;1/3 dx = 9a/8. (â&#x2C6;&#x2019;x1/3 ) â&#x2C6;&#x2019;a â&#x2C6;&#x2019;a

10. The base of the dome is a hexagon of side r. An equation of the circle of radius r that lies in a vertical x-y plane and passes through two opposite vertices of the base hexagon is x2 + y 2 = r2 . A horizontal, hexagonal cross section at height y above the base has area â&#x2C6;&#x161; â&#x2C6;&#x161; r â&#x2C6;&#x161; â&#x2C6;&#x161; 3 3 2 3 3 2 3 3 2 2 A(y) = (r â&#x2C6;&#x2019; y 2 ) dy = 3r3 . x = (r â&#x2C6;&#x2019; y ), hence the volume is V = 2 2 2 0

11. Let the sphere have radius R, the hole radius r. By the Pythagorean Theorem, r2 + (L/2)2 = R2 . Use cylindrical shells to â&#x2C6;&#x161; calculate the volume â&#x2C6;&#x161; of the solid obtained by rotating about the y-axis the region r < x < R, â&#x2C6;&#x2019; R2 â&#x2C6;&#x2019; x2 < y < R2 â&#x2C6;&#x2019; x2 : R R 4 4 (2Ď&#x20AC;x)2 R2 â&#x2C6;&#x2019; x2 dx = â&#x2C6;&#x2019; Ď&#x20AC;(R2 â&#x2C6;&#x2019; x2 )3/2 = Ď&#x20AC;(L/2)3 , V = 3 3 r r

L/2

12. V = 2

Ď&#x20AC; 0

16R2 2 4Ď&#x20AC; LR2 (x â&#x2C6;&#x2019; L2 /4)2 = 4 L 15

13. (a)

x 100

200

-0.4

(b) The maximum deďŹ&#x201A;ection occurs at x = 96 inches (the midpoint of the beam) and is about 1.42 in.

so the volume is independent of R.

-0.8 -1.2

-1.6

14. y = 0 at x = b = 30.585; distance =

1 + (12.54 â&#x2C6;&#x2019; 0.82x)2 dx = 196.306 yd

15. x = et (cos t â&#x2C6;&#x2019; sin t), y = et (cos t + sin t), (x )2 + (y )2 = 2e2t Ď&#x20AC;/2 â&#x2C6;&#x161; â&#x2C6;&#x161; Ď&#x20AC;/2 2t S = 2Ď&#x20AC; (et sin t) 2e2t dt = 2 2Ď&#x20AC; e sin t dt 0

uh a

â&#x2C6;&#x161;

Ď&#x20AC;/2 â&#x2C6;&#x161; 1 2 2 Ď&#x20AC;(2eĎ&#x20AC; + 1) = 2 2Ď&#x20AC; e2t (2 sin t â&#x2C6;&#x2019; cos t) = 5 5 0

16. (a) Ď&#x20AC;

(sinâ&#x2C6;&#x2019;1 x)2 dx = 1.468384.

Ď&#x20AC;/2

y(1 â&#x2C6;&#x2019; sin y)dy = 1.468384.

(b) 2Ď&#x20AC; 0

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Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Chapter 7 Supplementary Exercises

us uf

1/4 1 1 17. (a) F = kx, = k , k = 2, W = kx dx = 1/16 J 2 4 0 L kx dx = kL2 /2, L = 5 m (b) 25 = 0

150

(30x + 2000) dx = 15 · 1502 + 2000 · 150 = 637,500 lb·ft

18. F = 30x + 2000, W = 0

19. (a) F =

ρx3 dx N

w(x) x (b) By similar triangles = , w(x) = 2x, so 4 2 4 2 F = ρ(1 + x)2x dx lb/ft .

h(x) = 1 + x 0

w(x)

x 2

8 2 (c) A formula for the parabola is y = x − 10, so F = 125

20. y = a cosh ax, y = a2 sinh ax = a2 y

9810|y|2

−10

125 (y + 10) dy N. 8

21. (a) cosh 3x = cosh(2x + x) = cosh 2x cosh x + sinh 2x sinh x = (2 cosh2 x − 1) cosh x + (2 sinh x cosh x) sinh x = 2 cosh3 x − cosh x + 2 sinh2 x cosh x

(b) r = 1 when t ≈ 0.673080 s. (c) dr/dt = 4.48 m/s.

= 2 cosh3 x − cosh x + 2(cosh2 x − 1) cosh x = 4 cosh3 x − 3 cosh x x x (b) from Theorem 7.8.2 with x replaced by : cosh x = 2 cosh2 − 1, 2 2 1 2 x 2 x 2 cosh = cosh x + 1, cosh = (cosh x + 1), 2 2 2 x x 1 (cosh x + 1) (because cosh > 0) cosh = 2 2 2 x x (c) from Theorem 7.8.2 with x replaced by : cosh x = 2 sinh2 + 1, 2 2 1 x x x 1 (cosh x − 1) 2 sinh2 = cosh x − 1, sinh2 = (cosh x − 1), sinh = ± 2 2 2 2 2 22. (a)

t 1

uh a

315

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316

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Chapter 7

23. Set a = 68.7672, b = 0.0100333, c = 693.8597, d = 299.2239. (a)

(b) L = 2

650

1 + a2 b2 sinh2 bx dx

-300

300 0

(d) 82â&#x2014;Ś

us uf

= 1480.2798 ft

24. The x-coordinates of the points of intersection are a â&#x2030;&#x2C6; â&#x2C6;&#x2019;0.423028 and b â&#x2030;&#x2C6; 1.725171; the area is b (2 sin x â&#x2C6;&#x2019; x2 + 1)dx â&#x2030;&#x2C6; 2.542696.

25. Let (a, k), where Ď&#x20AC;/2 < a < Ď&#x20AC;, be the coordinates of the point of intersection of y = k with y = sin x. Thus k = sin a and if the shaded areas are equal, a a (k â&#x2C6;&#x2019; sin x)dx = (sin a â&#x2C6;&#x2019; sin x) dx = a sin a + cos a â&#x2C6;&#x2019; 1 = 0 0

Solve for a to get a â&#x2030;&#x2C6; 2.331122, so k = sin a â&#x2030;&#x2C6; 0.724611.

uh a

k = 1.736796.

x sin x dx = 2Ď&#x20AC;(sin k â&#x2C6;&#x2019; k cos k) = 8; solve for k to get

26. The volume is given by 2Ď&#x20AC;

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Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH CHAPTER 8

EXERCISE SET 8.1

1 1 1 u3 du = − u4 + C = − (3 − 2x)4 + C u = 3 − 2x, du = −2dx, − 2 8 8 1 2 3/2 2 u1/2 du = u = 4 + 9x, du = 9dx, u +C = (4 + 9x)3/2 + C 9 3·9 27 1 1 1 sec2 u du = tan u + C = tan(x2 ) + C u = x2 , du = 2xdx, 2 2 2 u = x2 , du = 2xdx, 2 tan u du = −2 ln | cos u | + C = −2 ln | cos(x2 )| + C

us uf

Principles of Integral Evaluation

1 dx, x

8. u = ln x, du =

du 1 1 1 5. u = 2 + cos 3x, du = −3 sin 3xdx, − = − ln |u| + C = − ln(2 + cos 3x) + C 3 u 3 3 du 3 2 du 1 1 1 3x , du = dx, = tan−1 u + C = tan−1 (3x/2) + C = 6. u = 6 6 1 + u2 6 2 2 3 4 + 4u2 sinh u du = cosh u + C = cosh ex + C 7. u = ex , du = ex dx,

sec u tan u du = sec u + C = sec(ln x) + C

9. u = cot x, du = − csc xdx,

eu du = −eu + C = −ecot x + C

−

du 1 1 = sin−1 u + C = sin−1 (x2 ) + C 2 2 1 − u2 1 1 1 u5 du = − u6 + C = − cos6 7x + C 11. u = cos 7x, du = −7 sin 7xdx, − 7 42 42 1 + 1 + sin2 x 1 + √1 + u2 du √ 12. u = sin x, du = cos x dx, = − ln + C = − ln +C u sin x u u2 + 1 1 2

√

10. u = x2 , du = 2xdx,

13. u = e , du = e dx, 14. u = tan−1 x, du = √

1 1 + x2

uh a 15. u =

du = ln u + u2 + 4 + C = ln ex + e2x + 4 + C 4 + u2 −1 dx, eu du = eu + C = etan x + C

√

1 x − 2, du = √ dx, 2 x−2

16. u = 3x2 + 2x, du = (6x + 2)dx,

17. u =

√

1 x, du = √ dx, 2 x

1 2

√ x−2

eu du = 2eu + C = 2e

cot u du =

1 1 ln | sin u| + C = ln sin |3x2 + 2x| + C 2 2

√ 2 cosh u du = 2 sinh u + C = 2 sinh x + C

317

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Chapter 8

2 2 , du = − 2 dx, x x

21. u =

−

1 2

csch2 u du =

1 1 2 coth u + C = coth + C 2 2 x

26. 27. 28.

25.

24.

23.

dx = ln x + x2 − 3 + C x2 − 3 1 2 + e−x du 1 2 + u −x −x + C = − ln +C = − ln u = e , du = −e dx, − 2 − u 4 2 − e−x 4 − u2 4 1 u = ln x, du = dx, cos u du = sin u + C = sin(ln x) + C x ex dx du x x √ √ = sin−1 u + C = sin−1 ex + C = u = e , du = e dx, 2x 1−e 1 − u2 1 − 2 sinh u du = −2 cosh u + C = −2 cosh(x−1/2 ) + C u = x−1/2 , du = − 3/2 dx, 2x du 1 1 1 1 2 cos u du = sin u + C = sin(x2 ) + C u = x , du = 2xdx, = 2 sec u 2 2 2 2du √ 2u = ex , 2du = ex dx, = sin−1 u + C = sin−1 (ex /2) + C 4 − 4u2 √

22.

us uf

du = ln |u| + C = ln |ln x| + C u √ 2 du 1 2 −u ln 3 2 −√x 19. u = x, du = √ dx, e 3 = 2 e−u ln 3 du = − +C =− +C u 3 ln 3 ln 3 2 x 20. u = sin θ, du = cos θdθ, sec u tan u du = sec u + C = sec(sin θ) + C

dx , 18. u = ln x, du = x

29. 4−x = e−x ln 4 , u = −x2 ln 4, du = −2x ln 4 dx = −x ln 16 dx, 1 u 1 −x2 ln 4 1 −x2 1 eu du = − e +C =− e 4 +C =− +C − ln 16 ln 16 ln 16 ln 16 1 πx ln 2 1 2πx dx = +C = 30. 2πx = eπx ln 2 , e 2πx + C π ln 2 π ln 2

EXERCISE SET 8.2 −x

1. u = x, dv = e

−x

dx, du = dx, v = −e

;

−x

dx = −xe

e−x dx = −xe−x − e−x + C

1 3x 1 1 1 e3x dx = xe3x − e3x + C xe − 3 3 3 9 3. u = x2 , dv = ex dx, du = 2x dx, v = ex ; x2 ex dx = x2 ex − 2 xex dx.

uh a

2. u = x, dv = e3x dx, du = dx, v =

1 3x e ; 3

xe3x dx =

xex dx use u = x, dv = ex dx, du = dx, v = ex to get

For

xe dx = xe − e + C1 so

318

x2 ex dx = x2 ex − 2xex + 2ex + C

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Exercise Set 8.2

−2x

4. u = x , dv = e

2 −2x

x e

1 dx = − x2 e−2x + 2

xe−2x dx

xe−2x dx use u = x, dv = e−2x dx to get

For

1 dx, du = 2x dx, v = − e−2x ; 2

319

us uf

1 1 1 1 xe−2x dx = − xe−2x + e−2x dx = − xe−2x − e−2x + C 2 2 2 4 1 1 1 so x2 e−2x dx = − x2 e−2x − xe−2x − e−2x + C 2 2 4

1 5. u = x, dv = sin 2x dx, du = dx, v = − cos 2x; 2 1 1 1 1 cos 2x dx = − x cos 2x + sin 2x + C x sin 2x dx = − x cos 2x + 2 4 2 2

x2 cos x dx = x2 sin x − 2

7. u = x2 , dv = cos x dx, du = 2x dx, v = sin x;

x sin x dx

x sin x dx use u = x, dv = sin x dx to get

For

1 6. u = x, dv = cos 3x dx, du = dx, v = sin 3x; 3 1 1 1 1 sin 3x dx = x sin 3x + cos 3x + C x cos 3x dx = x sin 3x − 3 3 3 9

x2 cos x dx = x2 sin x + 2x cos x − 2 sin x + C

x sin x dx = −x cos x + sin x + C1 so

8. u = x2 , dv = sin x dx, du = 2x dx, v = − cos x; 2 2 x sin x dx = −x cos x + 2 x cos x dx; for x cos x dx use u = x, dv = cos x dx to get

x cos x dx = x sin x + cos x + C1 so

x2 sin x dx = −x2 cos x + 2x sin x + 2 cos x + C

√ 1 2 9. u = ln x, dv = x dx, du = dx, v = x3/2 ; x 3 √ 2 3/2 2 2 4 x1/2 dx = x3/2 ln x − x3/2 + C x ln x dx = x ln x − 3 3 3 9 10. u = ln x, dv = x dx, du =

1 1 dx, v = x2 ; x 2

x ln x dx =

1 2 1 x ln x − 2 2

x dx =

1 2 1 x ln x − x2 + C 2 4

ln x dx, v = x; (ln x)2 dx = x(ln x)2 − 2 ln x dx. x Use u = ln x, dv = dx to get ln x dx = x ln x − dx = x ln x − x + C1 so

uh a

11. u = (ln x)2 , dv = dx, du = 2

(ln x)2 dx = x(ln x)2 − 2x ln x + 2x + C

√ 1 1 12. u = ln x, dv = √ dx, du = dx, v = 2 x; x x

√ ln x √ dx = 2 x ln x−2 x

√ √ 1 √ dx = 2 x ln x−4 x+C x

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Chapter 8

16. u = cos−1 (2x), dv = dx, du = − √

(2x)dx = x cos

−1

√

(2x) +

2x 1 dx = x cos−1 (2x) − 1 − 4x2 + C 2 1 − 4x2

cos

−1

2 dx, v = x; 1 − 4x2

us uf

2 2x 13. u = ln(2x + 3), dv = dx, du = dx, v = x; ln(2x + 3)dx = x ln(2x + 3) − dx 2x + 3 2x + 3 3 3 2x dx = 1− dx = x − ln(2x + 3) + C1 so but 2x + 3 2x + 3 2 3 ln(2x + 3)dx = x ln(2x + 3) − x + ln(2x + 3) + C 2 2x x2 14. u = ln(x2 + 4), dv = dx, du = 2 dx, v = x; ln(x2 + 4)dx = x ln(x2 + 4) − 2 dx x +4 x2 + 4 4 x x2 dx = x − 2 tan−1 + C1 so dx = 1− 2 but x +4 2 x2 + 4 x ln(x2 + 4)dx = x ln(x2 + 4) − 2x + 4 tan−1 + C 2 √ 15. u = sin−1 x, dv = dx, du = 1/ 1 − x2 dx, v = x; −1 −1 sin x dx = x sin x − x/ 1 − x2 dx = x sin−1 x + 1 − x2 + C

320

ex cos x dx use u = ex , dv = cos x dx to get

For

2 17. u = tan−1 (2x), dv = dx, du = dx, v = x; 1 + 4x2 2x 1 tan−1 (2x)dx = x tan−1 (2x) − dx = x tan−1 (2x) − ln(1 + 4x2 ) + C 1 + 4x2 4 x2 1 1 2 1 2 1 −1 −1 x x 18. u = tan−1 x, dv = x dx, du = dx, v = ; x tan x dx = tan x − dx 2 1+x 2 2 2 1 + x2 x2 1 but dx = x − tan−1 x + C1 so dx = 1− 1 + x2 1 + x2 1 1 1 x tan−1 x dx = x2 tan−1 x − x + tan−1 x + C 2 2 2 x x x x 19. u = e , dv = sin x dx, du = e dx, v = − cos x; e sin x dx = −e cos x + ex cos x dx.

ex sin x dx,

e sin x dx = e (sin x − cos x) + C1 ,

uh a

20. u = e2x , dv = cos 3x dx, du = 2e2x dx, v =

e2x cos 3x dx =

1 2x 2 e sin 3x − 3 3

ex sin x dx so

ex sin x dx = −ex cos x + ex sin x −

ex cos x = ex sin x −

ex sin x dx =

1 x e (sin x − cos x) + C 2

1 sin 3x; 3

e2x sin 3x dx. Use u = e2x , dv = sin 3x dx to get

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Exercise Set 8.2

1 2 sin 3x dx = − e2x cos 3x + 3 3

1 2 4 cos 3x dx = e2x sin 3x + e2x cos 3x − 3 9 9

e2x cos 3x dx so e2x cos 3x dx,

1 cos 3x dx = e2x (3 sin 3x + 2 cos 3x) + C1 , 9

e2x cos 3x dx =

13 9

us uf

1 2x e (3 sin 3x + 2 cos 3x) + C 13

1 21. u = eax , dv = sin bx dx, du = aeax dx, v = − cos bx (b = 0); b 1 a eax cos bx dx. Use u = eax , dv = cos bx dx to get eax sin bx dx = − eax cos bx + b b 1 a eax sin bx dx so eax cos bx dx = eax sin bx − b b 1 ax a ax a2 ax e sin bx dx = − e cos bx + 2 e sin bx − 2 eax sin bx dx, b b b eax eax sin bx dx = 2 (a sin bx − b cos bx) + C a + b2

e−3θ 22. From Exercise 21 with a = −3, b = 5, x = θ, answer = √ (−3 sin 5θ − 5 cos 5θ) + C 34 cos(ln x) dx, v = x; x sin(ln x)dx = x sin(ln x) − cos(ln x)dx. Use u = cos(ln x), dv = dx to get

23. u = sin(ln x), dv = dx, du =

cos(ln x)dx = x cos(ln x) +

sin(ln x)dx so

sin(ln x)dx = x sin(ln x) − x cos(ln x) −

sin(ln x)dx,

1 x[sin(ln x) − cos(ln x)] + C 2

sin(ln x)dx =

1 24. u = cos(ln x), dv = dx, du = − sin(ln x)dx, v = x; x cos(ln x)dx = x cos(ln x) + sin(ln x)dx. Use u = sin(ln x), dv = dx to get

uh a

sin(ln x)dx = x sin(ln x) −

cos(ln x)dx so

cos(ln x)dx = x cos(ln x) + x sin(ln x) −

cos(ln x)dx =

321

cos(ln x)dx,

1 x[cos(ln x) + sin(ln x)] + C 2

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Chapter 8

us uf

25. u = x, dv = sec2 x dx, du = dx, v = tan x; sin x 2 x sec x dx = x tan x − tan x dx = x tan x − dx = x tan x + ln | cos x| + C cos x 26. u = x, dv = tan2 x dx = (sec2 x − 1)dx, du = dx, v = tan x − x; x tan2 x dx = x tan x − x2 − (tan x − x)dx

x3 ex dx =

1 2 x2 x e − 2

xex dx =

1 x2 e ; 2

1 2 x2 1 x2 x e − e +C 2 2

1 1 = x tan x − x2 + ln | cos x| + x2 + C = x tan x − x2 + ln | cos x| + C 2 2 27. u = x2 , dv = xex dx, du = 2x dx, v =

1 1 ; dx, du = (x + 1)ex dx, v = − 2 (x + 1) x+1 xex xex xex ex x x + e + e +C dx = − dx = − + C = (x + 1)2 x+1 x+1 x+1

28. u = xex , dv =

1 29. u = x, dv = e−5x dx, du = dx, v = − e−5x ; 5 1 1 1 1 1 xe−5x dx = − xe−5x + e−5x dx 5 5 0 0 0 1 1 1 1 1 = − e−5 − e−5x = − e−5 − (e−5 − 1) = (1 − 6e−5 )/25 5 25 5 25 0

xe2x dx = 0

1 2x xe 2

2 − 0

1 2

1 2x e ; 2

30. u = x, dv = e2x dx, du = dx, v =

1 e2x dx = e4 − e2x 4

2 0

1 = e4 − (e4 − 1) = (3e4 + 1)/4 4

1 1 dx, v = x3 ; x 3 e e e e 1 1 1 1 1 1 x2 dx = e3 − x3 = e3 − (e3 − 1) = (2e3 + 1)/9 x2 ln x dx = x3 ln x − 3 3 3 9 3 9 1 1 1 1

31. u = ln x, dv = x2 dx, du =

1 1 1 dx, du = dx, v = − ; x2 x x e e ln x 1 1 ln x dx = − + dx 2 2 √ √ x x e x e e √ √ 1 1 3 e−4 1 1 1 1 1 =− + √ − +√ = = − + √ ln e − x √e e 2 e e 2e e e e

uh a

√ e

32. u = ln x, dv =

322

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Exercise Set 8.2

323

1 dx, v = x; x+3 2 2 2 2 3 x dx = 2 ln 5 + 2 ln 1 − dx 1− ln(x + 3)dx = x ln(x + 3) − x+3 −2 −2 x + 3 −2 −2 2 = 2 ln 5 − [x − 3 ln(x + 3)] = 2 ln 5 − (2 − 3 ln 5) + (−2 − 3 ln 1) = 5 ln 5 − 4

1 dx, v = x; 1 − x2 1/2 1/2 1/2 1/2 x 1 1 √ sin−1 x dx = x sin−1 x − dx = sin−1 + 1 − x2 2 2 1 − x2 0 0 0 0

√ 3 3 1 π π + = −1= + −1 2 6 4 12 2

√

34. u = sin−1 x, dv = dx, du = √

us uf

−2

1 √ dθ, v = θ; 2θ θ − 1 4 4 √ √ 4 1 4 √ √ 1 −1 −1 −1 −1 √ sec θdθ = θ sec θ − 2− θ−1 dθ = 4 sec 2 − 2 sec 2 2 θ−1 2 2 2 π √ π √ 5π −2 − 3+1= =4 − 3+1 6 3 4 θ, dv = dθ, du =

35. u = sec−1

1 1 36. u = sec−1 x, dv = x dx, du = √ dx, v = x2 ; 2 2 x x −1 2 2 x 1 2 1 2 −1 −1 √ x sec x dx = x sec x − dx 2 2 2 x −1 1 1 1 2 √ 1 2 1 x − 1 = 2π/3 − 3/2 = [(4)(π/3) − (1)(0)] − 2 2 1

1 37. u = x, dv = sin 4x dx, du = dx, v = − cos 4x; 4 π/2 π/2 π/2 1 1 π/2 1 x sin 4x dx = − x cos 4x + cos 4x dx = −π/8 + = −π/8 sin 4x 4 4 0 16 0 0 0

38.

(x + x cos x)dx =

1 2 x 2

x cos x dx = 0

π2 + 2

x cos x dx; 0

uh a

u = x, dv = cos x dx, du = dx, v = sin x π π π π π x cos x dx = x sin x − sin x dx = cos x = −2 so (x + x cos x)dx = π 2 /2 − 2 0

33. u = ln(x + 3), dv = dx, du =

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Chapter 8

√

2 1 dx, v = x3/2 ; xdx, du = √ 3 2 x(1 + x) 3 √ √ √ 3 1 3 x 2 dx x tan−1 xdx = x3/2 tan−1 x − 3 3 1 1+x 1 1 √ 3 1 3 2 1 = x3/2 tan−1 x − dx 1− 3 1 1+x 3 1 3 √ √ 2 3/2 1 1 = x tan−1 x − x + ln |1 + x| = (2 3π − π/2 − 2 + ln 2)/3 3 3 3 1 x, dv =

us uf

39. u = tan−1

324

2x dx, v = x; x2 + 1 2 2 2 2 1 2x2 2 2 dx dx = 2 ln 5 − 2 1− 2 ln(x + 1)dx = x ln(x + 1) − 2 x +1 0 x +1 0 0 0 2 −1 = 2 ln 5 − 2(x − tan x) = 2 ln 5 − 4 + 2 tan−1 2

40. u = ln(x2 + 1), dv = dx, du =

(a)

√ x

e (b)

cos cos

tet dt; u = t, dv = et dt, du = dt, v = et ,

dx = 2

dx = 2te − 2

√ √

√ √ et dt = 2(t − 1)et + C = 2( x − 1)e x + C

x, t2 = x, dx = 2t dt

x dx = 2

t cos t dt; u = t, dv = cos tdt, du = dt, v = sin t,

x dx = 2t sin t − 2

√

√ √ √ sin tdt = 2t sin t + 2 cos t + C = 2 x sin x + 2 cos x + C

41. t =

42. Let q1 (x), q2 (x), q3 (x) denote successive antiderivatives of q(x), so that q3 (x) = q2 (x), q2 (x) = q1 (x), q1 (x) = q(x). Let p(x) = ax2 + bx + c. Repeated Diﬀerentiation

Repeated Antidiﬀerentiation

ax2 + bx + c PP + PP PP 2ax + b PP P− PP P 2a PP P+ PP P 0

q(x) q1 (x) q2 (x) q3 (x)

p(x)q(x) dx = (ax2 + bx + c)q1 (x) − (2ax + b)q2 (x) + 2aq3 (x) + C. Check:

uh a

Then

d [(ax2 +bx + c)q1 (x) − (2ax + b)q2 (x) + 2aq3 (x)] dx

= (2ax + b)q1 (x) + (ax2 + bx + c)q(x) − 2aq2 (x) − (2ax + b)q1 (x) + 2aq2 (x) = p(x)q(x)

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Exercise Set 8.2

Repeated Antidiﬀerentiation e−x

3x2 − x + 2 PP 6x − 1 6

(3x2 − x + 2)e−x = −(3x2 − x + 2)e−x − (6x − 1)e−x − 6e−x + C = −e−x [3x2 + 5x + 7] + C

44.

Repeated Diﬀerentiation

Repeated Antidiﬀerentiation

P+ PP P −e−x PP P− PP P e−x PP P+ PP P −e−x

Repeated Diﬀerentiation

us uf

43.

325

x2 + x + 1 PP

sin x P+ PP P − cos x 2x + 1 PP P− PP P − sin x 2 PP P+ PP P cos x 0

(x2 + x + 1) sin x dx = −(x2 + x + 1) cos x + (2x + 1) sin x + 2 cos x + C

Repeated Antidiﬀerentiation

Repeated Diﬀerentiation

8x4 cos 2x PP + PP PP 1 32x3 sin 2x 2 PP − PP PP 1 − cos 2x 96x2 4 PP + PP PP 1 192x − sin 2x 8 PP − PP PP 1 192 cos 2x 16 PP + PP PP 1 0 sin 2x 32

uh a

45.

= −(x2 + x − 1) cos x + (2x + 1) sin x + C

8x4 cos 2x dx = (4x4 − 12x2 + 6) sin 2x + (8x3 − 12x) cos 2x + C

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Chapter 8

46.

Repeated Diﬀerentiation x3

Repeated Antidiﬀerentiation √ 2x + 1

❍

326

us uf

❍❍ + ❍❍ 1 3x (2x + 1)3/2 3 ❍❍ − ❍ ❍❍ 1 (2x + 1)5/2 6x 15 ❍ ❍+ ❍❍ ❍ 1 (2x + 1)7/2 6 105 ❍ ❍❍ − ❍❍ 1 0 (2x + 1)9/2 945

√ 1 1 2 2 x3 2x + 1 dx = x3 (2x + 1)3/2 − x2 (2x + 1)5/2 + x(2x + 1)7/2 − (2x + 1)9/2 + C 3 5 35 315 e ln x dx = (x ln x − x) = 1

47. (a) A = 1

(b) V = π

e (ln x)2 dx = π (x(ln x)2 − 2x ln x + 2x) = π(e − 2) 1

π/2

1 2 x 2

(x − x sin x)dx =

48. A =

π/2 π2 = π 2 /8 − 1 − (−x cos x + sin x) 8 0

π x sin x dx = 2π(−x cos x + sin x) = 2π 2

49. V = 2π 0

x sin x dx =

π/2

−

π/2

50. V = 2π

π/2 x cos x dx = 2π(cos x + x sin x) = π(π − 2)

51. distance =

t2 e−t dt; u = t2 , dv = e−t dt, du = 2tdt, v = −e−t ,

distance = −t2 e−t

te−t dt; u = 2t, dv = e−t dt, du = 2dt, v = −e−t ,

5 0

e−t dt = −25e−5 − 10e−5 − 2e−t

5 0

uh a

distance = −25e−5 − 2te−t

= −25e−5 − 10e−5 − 2e−5 + 2 = −37e−5 + 2

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Exercise Set 8.2

1 cos(kωt); the integrand is an even function of t so kω π/ω π/ω π/ω π/ω 1 2 t sin(kωt) dt = 2 t sin(kωt) dt = − t cos(kωt) +2 cos(kωt) dt kω kω −π/ω 0 0 0 π/ω 2 2π(−1)k+1 2π(−1)k+1 + sin(kωt) = = 2 2 2 kω k ω kω 2 0

327

us uf

52. u = 2t, dv = sin(kωt)dt, du = 2dt, v = −

1 2 sin x dx = − sin2 x cos x − cos x + C 3 3 1 3 1 1 3 2 4 sin x dx, sin2 x dx = − sin x cos x + x + C1 so (b) sin x dx = − sin x cos x + 2 2 4 4 π/4 π/4 1 3 3 sin4 x dx = − sin3 x cos x − sin x cos x + x 8 8 0 4 0 1 2 sin3 x dx = − sin2 x cos x + 3 3

53. (a)

√ √ √ √ 3 1 = − (1/ 2)3 (1/ 2) − (1/ 2)(1/ 2) + 3π/32 = 3π/32 − 1/4 4 8 1 4 1 4 1 2 5 4 3 4 2 cos x dx = cos x sin x + cos x sin x + sin x + C 54. (a) cos x dx = cos x sin x + 5 5 5 5 3 3 1 cos4 x sin x + 5 1 (b) cos6 x dx = cos5 x sin x + 6

4 8 cos2 x sin x + sin x + C 15 15 5 cos4 x dx 6 1 5 1 3 5 3 2 cos x dx = cos x sin x + cos x sin x + 6 4 4 6 1 1 5 5 1 cos3 x sin x + cos x sin x + x + C, = cos5 x sin x + 6 24 8 2 2 π/2 1 5 5 5 cos3 x sin x + cos x sin x + x cos5 x sin x + = 5π/32 24 16 16 0 6

55. u = sinn−1 x, dv = sin x dx, du = (n − 1) sinn−2 x cos x dx, v = − cos x; sinn x dx = − sinn−1 x cos x + (n − 1) sinn−2 x cos2 x dx n−1

sinn−2 x (1 − sin2 x)dx

x cos x + (n − 1)

= − sin

= − sinn−1 x cos x + (n − 1)

x cos x + (n − 1)

1 n−1 sinn−1 x cos x + n n

sinn−2 x dx,

sinn−2 x dx

uh a

sinn x dx,

n−1

sin x dx = − sin

sinn x dx = −

sinn−2 x dx − (n − 1)

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328

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Chapter 8

= secn−2 x tan x − (n − 2)

secn−2 x (sec2 x − 1)dx

= sec

n−2

x tan x − (n − 2)

sec x dx + (n − 2)

sec x dx = sec

n−2 1 secn−2 x tan x + n−1 n−1

secn−2 x dx

tan x dx =

(b)

n−2

tan

x (sec x − 1) dx =

1 = tann−1 x − n−1

tan

tann−2 x dx

n−1

57. (a)

1 tan3 x − 3

n−1

tann−2 x dx

n x

x e dx = x e − n

dx, v = e ;

tan2 x dx =

x sec x dx −

n x

secn x dx =

secn−2 x dx,

x tan x + (n − 2)

n−2

(n − 1)

secn−2 x dx,

us uf

56. (a) u = secn−2 x, dv = sec2 x dx, du = (n − 2) secn−2 x tan x dx, v = tan x; secn x dx = secn−2 x tan x − (n − 2) secn−2 x tan2 x dx

1 tan3 x − tan x + 3

dx =

xn−1 ex dx 1 tan3 x − tan x + x + C 3

1 2 1 2 2 sec2 x dx = sec2 x tan x + tan x + C (b) sec x dx = sec x tan x + 3 3 3 3 (c) x3 ex dx = x3 ex − 3 x2 ex dx = x3 ex − 3 x2 ex − 2 xex dx

x x = x e − 3x e + 6 xe − e dx = x3 ex − 3x2 ex + 6xex − 6ex + C 2 x

3 x

58. (a) u = 3x, 1 1 1 2 u 2 u2 eu du = u2 eu − 2 ueu du = ueu − eu du x2 e3x dx = u e − 27 27 27 27

√ (b) u = − x,

2 2 1 2 2 1 2 u u e − ueu + eu + C = x2 e3x − xe3x + e3x + C 27 27 3 9 27 27

ad =

√ x

xe−

−1

dx = 2

3 u

uh a

u e du = u e − 3

2 u u u e du = u e − 3 u e − 2 ue du 2 u

3 u

= u3 eu − 3u2 eu + 6 ueu − eu du = u3 eu − 3u2 eu + 6ueu − 6eu + C, −1

3 u

u e du = 2(u − 3u + 6u − 6)e

= 12 − 32e−1

−1

u3 eu du,

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Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Exercise Set 8.2

329

59. u = x, dv = f (x)dx, du = dx, v = f (x);

−1

1 x f (x)dx = xf (x) −

f (x)dx

−1

us uf

1 = f (1) + f (−1) − f (x) = f (1) + f (−1) − f (1) + f (−1)

−1

60. (a) u = f (x), dv = dx, du = f (x), v = x; b b b xf (x) dx = bf (b) − af (a) − f (x) dx = xf (x) −

xf (x) dx

f (a)

(b) Substitute y = f (x), dy = f (x) dx, x = a when y = f (a), x = b when y = f (b), b f (b) f (b) xf (x) dx = x dy = f −1 (y) dy f (a)

f (a)

which, by Part (b), yields β f −1 (x) dx = bf (b) − af (a) − −1

(β) − αf

(α) − β

Note from the ﬁgure that A1 =

f –1(a)

f –1(b)

f (x) dx

f −1 (β)

= βf

−1

a a=

bf (b) − af (a) = βf −1 (β) − αf −1 (α); then β β f (b) f −1 (x) dx = f −1 (y) dy = f −1 (y) dy, α

−1

f (x) dx

f −1 (α)

(x) dx, A2 =

f −1 (β)

f (x) dx, and f −1 (α)

A1 + A2 = βf −1 (β) − αf −1 (α), a ”picture proof”.

61. (a) Use Exercise 60(c); π/6 1/2 sin−1 (1/2) 1 1 1 1 −0·sin−1 0− − sin−1 x dx = sin−1 sin x dx = sin−1 sin x dx 2 2 2 2 −1 0 sin (0) 0

(b) Use Exercise 60(b); e2 ln x dx = e2 ln e2 − e ln e −

u dv = uv −

uh a

62. (a)

ln e2

ln e

f −1 (y) dy = 2e2 − e −

ey dy = 2e2 − e − 1

v du = x(sin x + C1 ) + cos x − C1 x + C2 = x sin x + cos x + C2 ;

the constant C1 cancels out and hence plays no role in the answer. (b) u(v + C1 ) − (v + C1 )du = uv + C1 u − v du − C1 u = uv − v du

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ex dx 1

Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Chapter 8

us uf

dx , v = x + 1; 63. u = ln(x + 1), dv = dx, du = x+1 ln(x + 1) dx = u dv = uv − v du = (x + 1) ln(x + 1) − dx = (x + 1) ln(x + 1) − x + C

1 (2x + 3) ln(2x + 3) − x + C 2

1 1 65. u = tan−1 x, dv = x dx, du = dx, v = (x2 + 1) 2 2 1+x 1 1 dx x tan−1 x dx = u dv = uv − v du = (x2 + 1) tan−1 x − 2 2

1 1 1 66. u = dx, v = ln x , dv = dx, du = − ln x x x(ln x)2 1 1 dx = 1 + dx. x ln x x ln x

1 2 1 (x + 1) tan−1 x − x + C 2 2

2dx 3 64. u = ln(2x + 3), dv = dx, du = ,v = x + ; 2x + 3 2 3 dx ln(2x + 3) dx = u dv = uv − v du = (x + ) ln(2x + 3) − 2 =

330

This seems to imply that 1 = 0, but recall that both sides represent a function plus an arbitrary constant; these two arbitrary constants will take care of the 1.

1 u du = − cos6 x + C 6 5

1. u = cos x, −

1 cos 3x dx = 2 1 sin 5θ dθ = 2 2

cos at dt =

uh a

u4 du =

1 sin5 3x + C 15

(1 + cos 6x)dx =

1 1 x+ sin 6x + C 2 12

(1 − cos 10θ)dθ =

1 1 θ− sin 10θ + C 2 20

1 2. u = sin 3x, 3

1 sin2 ax + C, a = 0 2a

(1 − sin2 at) cos at dt

u du =

1 3. u = sin ax, a

EXERCISE SET 8.3

cos θdθ =

cos at dt −

sin2 at cos at dt =

1 1 sin at − sin3 at + C (a = 0) a 3a

(1 − sin θ) cos θdθ =

= sin θ −

(1 − 2 sin2 θ + sin4 θ) cos θdθ

1 2 sin3 θ + sin5 θ + C 3 5

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Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Exercise Set 8.3

sin3 x(1 − sin2 x) cos x dx

sin x cos x dx =

(sin3 x − sin5 x) cos x dx =

sin2 2t cos3 2t dt =

sin2 2t(1 − sin2 2t) cos 2t dt =

(1 − cos2 2x) cos2 2x sin 2x dx

sin 2x cos 2x dx =

1 1 cos5 2x + C (cos2 2x − cos4 2x) sin 2x dx = − cos3 2x + 10 6

sin2 x cos2 x dx =

1 4

sin2 x cos4 x dx =

1 8

1 = 8

13.

sin 3θ cos 2θdθ =

15.

16. u = cos x, −

(1 − cos 2x)(1 + cos 2x)2 dx =

cos x dx =

1 2

(sin 5θ + sin θ)dθ = −

1 sin3 2x 48

1 1 cos 5θ − cos θ + C 10 2

π/4 √ √ 1 1 √ = sin x − sin3 x = ( 2/2) − ( 2/2)3 = 5 2/12 3 3 0 1 4

π/3

1 = 8

sin4 3x cos3 3x dx =

(1 − cos 4x)dx +

(1 − sin2 x) cos x dx

π/2

uh a M

1 [sin(3x/2) + sin(x/2)]dx = − cos(3x/2) − cos(x/2) + C 3

19.

1 sin 2x cos 2x dx = 16 2

1 1 (sin 3x − sin x)dx = − cos 3x + cos x + C 6 2

sin2 (x/2) cos2 (x/2)dx =

18.

(1 − cos2 2x)(1 + cos 2x)dx

π/4

17.

1 sin 2x dx + 8 2

1 8

1 1 x− sin 4x + C 8 32

5 u1/5 du = − cos6/5 x + C 6

π/4 3

1 sin x cos(x/2)dx = 2

(1 − cos 4x)dx =

1 1 1 x− sin 4x + sin3 2x + C 16 64 48

1 sin x cos 2x dx = 2

14.

1 8

sin2 2x dx =

12.

11.

10.

(sin2 2t − sin4 2t) cos 2t dt

1 1 sin3 2t − sin5 2t + C 6 10

1 1 sin4 x − sin6 x + C 4 6

us uf

331

π/2

sin2 x dx = 0

1 8

π/2

(1 − cos 2x)dx 0

π/2 1 x − sin 2x = π/16 2 0

π/3

sin4 3x(1 − sin2 3x) cos 3x dx = 0

π/3 1 1 =0 sin5 3x − sin7 3x 15 21 0

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Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Chapter 8

cos2 5θ dθ = −π

π/6

1 2

(1 + cos 10θ)dθ = −π

1 sin 2x cos 4x dx = 2

21. 0

π/6

1 2

θ+

1 sin 10θ 10

π =π −π

π/6 1 1 (sin 6x − sin 2x)dx = − cos 6x + cos 2x 12 4 0

20.

us uf

332

= [(−1/12)(−1) + (1/4)(1/2)] − [−1/12 + 1/4] = 1/24

1 (1 − cos 2kx)dx = 2

25. u = e

−2x

, du = −2e

1 dx; − 2

tan u du =

27.

u2 du =

√ √ 2 sec u du = 2 ln | sec u + tan u| + C = 2 ln sec x + tan x + C

1 tan3 x + C 3

tan5 x(1 + tan2 x) sec2 x dx =

30.

(tan5 x + tan7 x) sec2 x dx =

(sec2 θ − 1)2 sec θ tan θdθ =

uh a

1 3 sec3 x tan x + 4 4

(sec4 θ − 2 sec2 θ + 1) sec θ tan θdθ =

(sec x − 1) sec x dx = =

(sec6 x − sec4 x) sec x tan x dx =

34.

1 1 tan4 4x + tan6 4x + C 16 24

sec4 x(sec2 x − 1) sec x tan x dx =

33.

(tan3 4x + tan5 4x) sec2 4x dx =

1 1 tan5 θ + tan7 θ + C 5 7

tan4 θ(1 + tan2 θ) sec2 θ dθ =

32.

1 1 tan6 x + tan8 x + C 6 8

tan3 4x(1 + tan2 4x) sec2 4x dx =

31.

1 ln | sec 2x + tan 2x| + C 2

1 x, du = √ dx; 2 x

29. u = tan x,

1 1 ln | cos u| + C = ln | cos(e−2x )| + C 2 2

√

2π 1 1 x− =π− sin 2kx sin 4πk (k = 0) 2k 4k 0

1 24. − ln | cos 5x| + C 5

1 ln | sin 3x| + C 3

28. u =

35.

1 tan(3x + 1) + C 3 −2x

26.

2π

22.

23.

1 sin kx dx = 2 2

2π

1 1 sec7 x − sec5 x + C 7 5

1 2 sec5 θ − sec3 θ + sec θ + C 5 3

(sec x − 2 sec x + sec x)dx =

sec x dx − 2

sec x dx +

sec3 x dx − 2

sec3 x dx + ln | sec x + tan x|

1 5 1 1 3 = sec x tan x − sec x tan x + ln | sec x + tan x| + ln | sec x + tan x| + C 4 4 2 2 =

5 3 1 sec3 x tan x − sec x tan x + ln | sec x + tan x| + C 4 8 8

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sec x dx

Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Exercise Set 8.3

[sec (x/2) − 1] sec (x/2)dx = [sec5 (x/2) − sec3 (x/2)]dx 5 3 =2 sec u du − sec u du

(u = x/2)

us uf

36.

1 3 3 3 3 sec u du − sec u du =2 sec u tan u + 4 4 1 1 3 sec3 u du = sec u tan u − 2 2

(equation (20))

sec2 2t(sec 2t tan 2t)dt = sec4 x dx =

38.

(1 + tan2 x) sec2 x dx =

1 sec5 x + C 5

(sec2 x + tan2 x sec2 x)dx = tan x +

1 tan3 x + C 3

40. Using equation (20), 1 3 sec5 x dx = sec3 x tan x + sec3 x dx 4 4 =

sec4 x(sec x tan x)dx =

39.

1 sec3 2t + C 6

37.

(equation (20), (22))

1 1 1 sec3 u tan u − sec u tan u − ln | sec u + tan u| + C 2 4 4 x x 1 x 1 x 1 x x = sec3 tan − sec tan − ln sec + tan + C 2 2 2 4 2 2 4 2 2

3 1 3 sec3 x tan x + sec x tan x + ln | sec x + tan x| + C 8 8 4

tan4 x dx =

1 tan3 x − tan x + x + C 3

41. Use equation (19) to get

42. u = 4x, use equation (19) to get 1 1 1 1 1 3 2 tan u du = tan u + ln | cos u| + C = tan2 4x + ln | cos 4x| + C 4 4 2 8 4 √

tan x(1 + tan2 x) sec2 x dx =

43.

sec1/2 x(sec x tan x)dx =

44.

π/6

(sec2 2x − 1)dx =

45. 0

2 2 tan3/2 x + tan7/2 x + C 3 7

2 sec3/2 x + C 3

π/6 √ 1 tan 2x − x = 3/2 − π/6 2 0

1 sec θ(sec θ tan θ)dθ = sec3 θ 3

uh a

π/6

46.

√ √ = (1/3)(2/ 3)3 − 1/3 = 8 3/27 − 1/3

47. u = x/2,

333

π/4

tan5 u du =

2 0

π/4 √ 1 = 1/2 − 1 − 2 ln(1/ 2) = −1/2 + ln 2 tan4 u − tan2 u − 2 ln | cos u| 2 0

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Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Chapter 8

1 sec u tan u du = sec u π

(csc2 x − 1) csc2 x(csc x cot x)dx = 50.

1 cos2 3t dt = · 2 sin 3t cos 3t

1 1 (csc4 x − csc2 x)(csc x cot x)dx = − csc5 x + csc3 x + C 5 3

(csc2 x − 1) cot x dx =

52.

1 csc 3t cot 3t dt = − csc 3t + C 3

51.

√ = ( 2 − 1)/π

csc x(csc x cot x)dx −

cos x 1 dx = − csc2 x − ln | sin x| + C sin x 2

1 (cot2 x + 1) csc2 x dx = − cot3 x − cot x + C 3

2π

53. (a)

2π cos(m + n)x cos(m − n)x − [sin(m+n)x+sin(m−n)x]dx = − 2(m + n) 2(m − n) 0 0 2π = 0, cos(m − n)x = 0. 1 2

sin mx cos nx dx = 0

2π but cos(m + n)x

2π

49.

π/4

us uf

1 48. u = πx, π

334

2π

1 2π [cos(m + n)x + cos(m − n)x]dx; 2 0 0 since m = n, evaluate sin at integer multiples of 2π to get 0. 2π 1 2π sin mx sin nx dx = [cos(m − n)x − cos(m + n)x] dx; (c) 2 0 0 since m = n, evaluate sin at integer multiples of 2π to get 0. 2π

54. (a) 0

2π

(b) 0

1 sin mx cos mx dx = 2 1 cos2 mx dx = 2

2π

2π 1 sin 2mx dx = − cos 2mx = 0 4m

1 (1 + cos 2mx) dx = 2

cos mx cos nx dx =

(b)

2π 1 x+ sin 2mx = π 2m 0

2π 2π 1 1 2π 1 2 x− sin mx dx = (1 − cos 2mx) dx = (c) sin 2mx = π 2 2 2m 0 0 0

55. y = tan x, 1 + (y )2 = 1 + tan2 x = sec2 x,

√

uh a

π/4

56. V = π

sec2 x dx =

π/4

π/4 √ sec x dx = ln | sec x + tan x| = ln( 2 + 1)

π/4

(1 − tan2 x)dx = π 0

π/4

π/4 1 (2 − sec2 x)dx = π(2x − tan x) = π(π − 2) 2 0

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Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Exercise Set 8.3

π/4

(cos x − sin x)dx = π

57. V = π 0

π sin x dx = 2

58. V = π 0

π/4 1 cos 2x dx = π sin 2x = π/2 2 0

π (1 − cos 2x)dx = 2

L 59. With 0 < α < β, D = Dβ −Dα = 2π

π 1 x − sin 2x = π 2 /2 2 0

L sec x dx = ln | sec x + tan x| 2π

100 ln(sec 25◦ + tan 25◦ ) = 7.18 cm 2π 100 sec 50◦ + tan 50◦ = 7.34 cm (b) D = ln sec 30◦ + tan 30◦ 2π csc x dx =

= α

L sec β + tan β ln sec α + tan α 2π

sec(π/2 − x)dx = − ln | sec(π/2 − x) + tan(π/2 − x)| + C

61. (a)

60. (a) D =

us uf

π/4

335

= − ln | csc x + cot x| + C

(b) − ln | csc x + cot x| = ln

√

2 [sin x cos(π/4) + cos x sin(π/4)] =

dx 1 =√ sin x + cos x 2

√

2 sin(x + π/4),

1 csc(x + π/4)dx = − √ ln | csc(x + π/4) + cot(x + π/4)| + C 2 √ 2 + cos x − sin x 1 = − √ ln +C sin x + cos x 2

√ √ √ 2 (1/ 2) sin x + (1/ 2) cos x

62. sin x + cos x =

a b a2 + b2 √ sin x + √ cos x = a2 + b2 (sin x cos θ + cos x sin θ) a2 + b2 a2 + b2 √ √ √ where cos θ = a/ a2 + b2 and sin θ = b/ a2 + b2 so a sin x + b cos x = a2 + b2 sin(x + θ)

uh a

63. a sin x + b cos x =

and

1 ln | csc(x + θ) + cot(x + θ)| + C + b2 √ a2 + b2 + a cos x − b sin x 1 = −√ ln +C 2 2 a sin x + b cos x a +b

dx 1 =√ 2 a sin x + b cos x a + b2

csc(x + θ)dx = − √

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Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Chapter 8

π/2

π/2 1 n − 1 π/2 n−2 n − 1 π/2 n−2 n−1 sin x dx = − sin x cos x + sin x dx = sin x dx n n n 0 0 0 n

64. (a) 0

us uf

(b) By repeated application of the formula in Part (a) π/2 π/2 n−3 n−1 n sin x dx = sinn−4 x dx n n−2 0 0  π/2  n−3 n−5 1 n−1   · · · dx, n even   n n−2 n−4 2 0 = π/2   n−3 n−5 2 n−1   ··· sin x dx, n odd  n n−2 n−4 3 0  1 · 3 · 5 · · · (n − 1) π   · , n even  2 · 4 · 6···n 2 =    2 · 4 · 6 · · · (n − 1) , n odd 3 · 5 · 7···n π/2 π/2 2 1·3 π 65. (a) sin3 x dx = sin4 x dx = (b) · = 3π/16 3 2·4 2 0 0 π/2 π/2 2·4 1·3·5 π sin5 x dx = sin6 x dx = (c) = 8/15 (d) · = 5π/32 3 · 5 2 ·4·6 2 0 0

336

66. Similar to proof in Exercise 64.

EXERCISE SET 8.4

1. x = 2 sin θ, dx = 2 cos θ dθ, 4 cos2 θ dθ = 2 (1 + cos 2θ)dθ = 2θ + sin 2θ + C

1 = 2θ + 2 sin θ cos θ + C = 2 sin−1 (x/2) + x 4 − x2 + C 2 1 1 sin θ, dx = cos θ dθ, 2 2 1 1 1 1 cos2 θ dθ = (1 + cos 2θ)dθ = θ + sin 2θ + C 2 4 4 8

2. x =

1 1 1 1 θ + sin θ cos θ + C = sin−1 2x + x 1 − 4x2 + C 4 4 4 2

uh a

3. x = 3 sin θ, dx = 3 cos θ dθ, 9 9 9 9 9 (1 − cos 2θ)dθ = θ − sin 2θ + C = θ − sin θ cos θ + C 9 sin2 θ dθ = 2 2 4 2 2 9 1 = sin−1 (x/3) − x 9 − x2 + C 2 2

4. x = 4 sin θ, dx = 4 cos θ dθ, √ 1 16 − x2 1 1 1 2 csc θ dθ = − cot θ + C = − +C dθ = 2 16 16 16x 16 sin θ

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Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Exercise Set 8.4

x 1 1 1 x +C θ+ sin θ cos θ + C = tan−1 + 16 16 16 2 8(4 + x2 )

us uf

5. x = 2 tan θ, dx = 2 sec2 θ dθ, 1 1 1 1 1 1 2 dθ = cos (1 + cos 2θ)dθ = θ+ sin 2θ + C θ dθ = 2 8 sec θ 8 16 16 32

337

√ √ 6. x = 5 tan θ, dx = 5 sec2 θ dθ, 1 1 2 3 sec θ tan θ − ln | sec θ + tan θ| + C1 5 tan θ sec θ dθ = 5 (sec θ − sec θ)dθ = 5 2 2 √ 5 + x2 + x 1 5 5 1 √ = x 5 + x2 − ln + C1 = x 5 + x2 − ln( 5 + x2 + x) + C 2 2 2 2 5

7. x = 3 sec θ, dx = 3 sec θ tan θ dθ, x 3 tan2 θ dθ = 3 (sec2 θ − 1)dθ = 3 tan θ − 3θ + C = x2 − 9 − 3 sec−1 + C 3

√

2 cos θ dθ, √ √ 1 − cos2 θ sin θ dθ 2 2 sin3 θ dθ = 2 2

9. x =

2 sin θ, dx =

8. x = 4 sec θ, dx = 4 sec θ tan θ dθ, √ 1 1 1 x2 − 16 1 cos θ dθ = dθ = sin θ + C = +C sec θ 16 16 16x 16

√ 1 1 3 = 2 2 − cos θ + cos θ + C = −2 2 − x2 + (2 − x2 )3/2 + C 3 3 √

√ 5 sin θ, dx = 5 cos θ dθ, √ √ 1 1 5 1 25 5 sin3 θ cos2 θ dθ = 25 5 − cos3 θ + cos5 θ + C = − (5 − x2 )3/2 + (5 − x2 )5/2 + C 3 5 3 5

10. x =

3 2 3 11. x = sec θ, dx = sec θ tan θ dθ, 2 2 9

1 2 dθ = sec θ 9

2 cos θ dθ = sin θ + C = 9

√

4x2 − 9 +C 9x

12. t = tan θ, dt = sec2 θ dθ, sec3 θ tan2 θ + 1 dθ = sec θ dθ = (sec θ tan θ + csc θ)dθ tan θ tan θ √ 1 + t2 − 1 2 = sec θ + ln | csc θ − cot θ| + C = 1 + t + ln +C |t|

uh a

13. x = sin θ, dx = cos θ dθ,

1 dθ = cos2 θ

1 14. x = 5 tan θ, dx = 5 sec θ dθ, 25 2

15. x = sec θ, dx = sec θ tan θ dθ,

sec2 θ dθ = tan θ + C = x/ 1 − x2 + C

√ sec θ 1 1 x2 + 25 +C 2 dθ = 25 csc θ cot θ dθ = − 25 csc θ+C = − 25x tan θ

sec θ dθ = ln | sec θ + tan θ| + C = ln x + x2 − 1 + C

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338

Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Chapter 8

1 1 1 x θ + sin θ cos θ + C = tan−1 x + +C 2 2 2 2(1 + x2 )

1 1 sec θ, dx = sec θ tan θ dθ, 3 3 sec θ 1 1 1 csc θ cot θ dθ = − csc θ + C = −x/ dθ = 9x2 − 1 + C 3 3 3 tan2 θ

us uf

16. 1 + 2x2 + x4 = (1 + x2 )2 , x = tan θ, dx = sec2 θ dθ, 1 1 1 1 2 dθ = cos (1 + cos 2θ)dθ = θ + sin 2θ + C θ dθ = sec2 θ 2 2 4

18. x = 5 sec θ, dx = 5 sec θ tan θ dθ, 25 25 sec θ tan θ + ln | sec θ + tan θ| + C1 25 sec3 θ dθ = 2 2 25 1 = x x2 − 25 + ln |x + x2 − 25| + C 2 2

17. x =

20. u = sin θ,

1 √ du = sin−1 2 − u2

sin θ √ 2

19. ex = sin θ, ex dx = cos θ dθ, 1 1 1 1 1 2 cos θ dθ = (1 + cos 2θ)dθ = θ + sin 2θ + C = sin−1 (ex ) + ex 1 − e2x + C 2 4 2 2 2

21. x = 4 sin θ, dx = 4 cos θ dθ, π/2 π/2 1 1 3 2 3 5 1024 sin θ cos θ dθ = 1024 − cos θ + cos θ = 1024(1/3 − 1/5) = 2048/15 3 5 0 0 2 2 sin θ, dx = cos θ dθ, 3 3 π/6 π/6 π/6 1 1 1 1 1 3 sec θ dθ = dθ = sec θ tan θ + ln | sec θ + tan θ| 24 0 cos3 θ 24 0 48 48 0 √ √ √ √ 1 2 1 1 [(2/ 3)(1/ 3) + ln |2/ 3 + 1/ 3|] = = + ln 3 48 48 3 2

22. x =

π/3

√

23. x = sec θ, dx = sec θ tan θ dθ,

24. x =

2 sec θ, dx =

√

π/4

1 dθ = sec θ

π/4

π/3

cos θ dθ = sin θ π/4

√ √ = ( 3 − 2)/2

π/4

π/4 tan θ dθ = 2 tan θ − 2θ = 2 − π/2 2

2 sec θ tan θ dθ, 2 0

uh a

√ √ 25. x = 3 tan θ, dx = 3 sec2 θ dθ, √3/2 1 π/3 cos3 θ 1 − u2 1 π/3 sec θ 1 π/3 1 − sin2 θ 1 dθ = du (u = sin θ) dθ = cos θ dθ = 9 π/6 tan4 θ 9 π/6 sin4 θ 9 π/6 9 1/2 u4 sin4 θ √ √3/2 √3/2 1 1 1 10 3 + 18 1 −4 −2 − 3+ (u − u )du = = = 9 3u u 1/2 243 9 1/2

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√

tan θ dθ = ln | sec θ| + C1 = ln

1 ln(x2 + 4) + C with C = C1 − ln 2 2

2 tan2 θ dθ = 2 tan θ − 2θ + C = x − 2 tan−1

x2 dx = 2 x +4

x + C; alternatively 2

28. x = 2 tan θ, dx = 2 sec2 θ dθ,

x2 + 4 + C1 = ln(x2 + 4)1/2 − ln 2 + C1 2

27. u = x2 + 4, du = 2x dx, 1 1 1 1 du = ln |u| + C = ln(x2 + 4) + C; or x = 2 tan θ, dx = 2 sec2 θ dθ, 2 u 2 2

us uf

√ √ 26. x = 3 tan θ, dx = 3 sec2 θ dθ, √ π/3 √ π/3 √ π/3 3 tan3 θ 3 3 3 dθ = 1 − cos2 θ sin θ dθ sin θ dθ = 3 0 sec3 θ 3 0 3 0 √ √ π/3 √ 1 1 1 1 3 3 3 − cos θ + cos θ − + − −1 + = 5 3/72 = = 3 3 3 2 24 3 0

dx x = x − 2 tan−1 + C +4 2

dx − 4

1 1 x2 + 1 , , 1 + (y )2 = 1 + 2 = x x x2 2√ 2 2 2 x +1 x +1 dx = L= dx; x = tan θ, dx = sec2 θ dθ, 2 x x 1 1

tan−1 (2) tan2 θ + 1 L= (sec θ tan θ + csc θ)dθ sec θ dθ = tan θ π/4 π/4 π/4 √ tan−1 (2) √

√ √ 5 1 − = sec θ + ln | csc θ − cot θ| = 5 + ln 2 + ln | 2 − 1| − 2 2 π/4 √ √ √ 2+2 2 √ = 5 − 2 + ln 1+ 5 tan−1 (2)

sec3 θ dθ = tan θ

tan−1 (2)

29. y =

30. y = 2x, 1 + (y )2 = 1 + 4x2 , 1

1 1 1 + 4x2 dx; x = tan θ, dx = sec2 θ dθ, 2 2

1 2

tan−1 2

sec3 θ dθ =

1 2

tan−1 2 1 1 sec θ tan θ + ln | sec θ + tan θ| 2 2 0

uh a

√ √ 1 √ 1 1√ 1 = ( 5)(2) + ln | 5 + 2| = 5 + ln(2 + 5) 4 4 2 4

31. y = 2x, 1 + (y )2 = 1 + 4x2 ,

S = 2π 0

1 + 4x2 dx; x =

339

Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Exercise Set 8.4

1 1 tan θ, dx = sec2 θ dθ, 2 2

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Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Chapter 8

tan−1 2

π tan θ sec θ dθ = 4 2

tan−1 2

π (sec θ − 1) sec θ dθ = 4 2

tan−1 2

(sec5 θ − sec3 θ)dθ 0

tan−1 2 √ √ 1 π 1 1 π sec3 θ tan θ − sec θ tan θ − ln | sec θ + tan θ| [18 5 − ln(2 + 5)] = 8 8 32 4 4 0

32. V = π

1 − y 2 dy; y = sin θ, dy = cos θ dθ,

0 π/2

π sin θ cos θ dθ = 4 2

V =π 0

π/2

π sin 2θ dθ = 8

π/2

π (1 − cos 4θ)dθ = 8

du = u + C = sinh−1 (x/3) + C

33. (a) x = 3 sinh u, dx = 3 cosh u du,

1 θ − sin 4θ 4

π/2 π2 = 16 0

(b) x = 3 tan θ, dx = 3 sec2 θ dθ, sec θ dθ = ln | sec θ + tan θ| + C = ln x2 + 9/3 + x/3 + C

√ but sinh−1 (x/3) = ln x/3 + x2 /9 + 1 = ln x/3 + x2 + 9/3 so the results agree.

a2 − x2 dx; x = a cos θ, dx = −a sin θ dθ,

38.

1 − (x − 1)2

(1 − cos 2θ) dθ = πab 0

x−2 3

9 − (x − 1)2

dx = sin−1 (x − 1) + C

−1

(x − 3)2 + 1

dx = sin

1 1 dx = 16(x + 1/2)2 + 1 16

39.

π/2

sin θ dθ = 2ab

π/2

a sin θ dθ = 4ab

uh a

π/2

37.

1 1 dx = tan−1 (x − 2)2 + 9 3

36.

35.

4b A=− a

(c) x = cosh u, dx = sinh u du, 1 1 1 2 (cosh 2u − 1)du = sinh 2u − u + C sinh u du = 4 2 2 1 1 1 1 = sinh u cosh u − u + C = x x2 − 1 − cosh−1 x + C 2 2 2 2 √ because cosh u = x, and sinh u = cosh2 u − 1 = x2 − 1 4b 34. A = a

x−1 3

π S= 4

us uf

340

1 1 dx = tan−1 (4x + 2) + C (x + 1/2)2 + 1/16 4

dx = ln x − 3 + (x − 3)2 + 1 + C

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Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Exercise Set 8.4

341

x dx, let u = x + 3, (x + 3)2 + 1 1 u−3 u 3 du = ln(u2 + 1) − 3 tan−1 u + C du = − 2 2 2 u +1 u +1 u +1 2

us uf

40.

1 ln(x2 + 6x + 10) − 3 tan−1 (x + 3) + C 2

cos2 θ dθ = 2θ + sin 2θ + C = 2θ + 2 sin θ cos θ + C

= 2 sin−1

x+1 2

1 + (x + 1) 3 − 2x − x2 + C 2

ex dx, let u = ex + 1/2, (ex + 1/2)2 + 3/4

42.

4 − (x + 1)2 dx, let x + 1 = 2 sin θ, 41.

x √ 1 2e + 1 √ du = sinh−1 (2u/ 3) + C = sinh−1 +C 3 u2 + 3/4 √ 3 x Alternate solution: let e + 1/2 = tan θ, 2 √ 2 e2x + ex + 1 2ex + 1 √ sec θ dθ = ln | sec θ + tan θ| + C = ln + √ + C1 3 3 = ln(2 e2x + ex + 1 + 2ex + 1) + C

1 1 dx = 2 2(x + 1) + 5 2

2x + 3 dx, let u = x + 1/2, 4(x + 1/2)2 + 4 u 1 2u + 2 1 1 1 du = ln(u2 + 1) + tan−1 u + C du = + 2 2 2 4u + 4 2 u +1 u +1 4 2 = 2

45.

46. 0

−π/6

−π/2

4x − x2 dx =

4 − (x − 2)2

−1

dx = sin

x−2 2

2 = π/6 1

4 − (x − 2)2 dx, let x − 2 = 2 sin θ,

√ −π/6 3 2π cos θ dθ = 2θ + sin 2θ = − 3 2 −π/2

uh a

1 √ dx = 4x − x2

1 1 ln(x2 + x + 5/4) + tan−1 (x + 1/2) + C 4 2

44.

1 1 dx = √ tan−1 2/5(x + 1) + C 2 (x + 1) + 5/2 10

43.

47. u = sin2 x, du = 2 sin x cos x dx;

1 1 2 1 u 1 − u2 + sin−1 u + C = sin x 1 − sin4 x + sin−1 (sin2 x) + C 1 − u2 du = 4 4 2

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342

Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Chapter 8

us uf

48. u = x sin x, du = (x cos x + sin x) dx; 1 1 1 1 1 + u2 du = u 1 + u2 + sinh−1 u + C = x sin x 1 + x2 sin2 x + sinh−1 (x sin x) + C 2 2 2 2

EXERCISE SET 8.5 A B + (x − 2) (x + 5)

B C A 5 + = + x x−3 x+3 x(x − 3)(x + 3)

2x − 3 C A B = + 2+ x−1 x2 (x − 1) x x

C A B + + (x + 2)3 x + 2 (x + 2)2

C Dx + E A B + 2+ 3+ 2 x x x x +1

A Bx + C + 2 x−1 x +5

Cx + D Ax + B + 2 2 x +5 (x + 5)2

Bx + C Dx + E A + 2 + 2 x−2 x +1 (x + 1)2

12.

A B 5x − 5 = + ; A = 1, B = 2 so (x − 3)(3x + 1) x − 3 3x + 1 1 2 1 dx + 2 dx = ln |x − 3| + ln |3x + 1| + C x−3 3x + 1 3 2x2 − 9x − 9 A B C = + + ; A = 1, B = 2, C = −1 so x(x + 3)(x − 3) x x+3 x−3 x(x + 3)2 1 1 1 +C dx + 2 dx − dx = ln |x| + 2 ln |x + 3| − ln |x − 3| + C = ln x x+3 x−3 x−3

uh a

13.

11x + 17 A B = + ; A = 5, B = 3 so (2x − 1)(x + 4) 2x − 1 x + 4 1 5 1 dx + 3 dx = ln |2x − 1| + 3 ln |x + 4| + C 5 x+4 2 2x − 1

11.

B 1 1 A 1 + ; A = , B = − so = x+1 x+7 6 6 (x + 1)(x + 7) 1 1 1 1 1 1 1 x + 1 +C dx − dx = ln |x + 1| − ln |x + 7| + C = ln 6 x+1 6 x+7 6 6 6 x + 7

10.

1 A B 1 1 = + ; A = − , B = so (x + 4)(x − 1) x+4 x−1 5 5 1 1 1 1 1 1 x − 1 1 dx + dx = − ln |x + 4| + ln |x − 1| + C = ln +C − 5 x+4 5 x−1 5 5 5 x + 4

Note that the symbol C has been recycled; to save space this recycling is usually not mentioned.

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15.

6 x2 + 2 , =x−2+ x+2 x+2

16.

x2 − 4 3 =x+1− , x−1 x−1

6 x−2+ x+2

x+1−

3 x−1

dx =

1 2 x − 2x + 6 ln |x + 2| + C 2

dx =

1 2 x + x − 3 ln |x − 1| + C 2

B 3x2 − 10 A 12x − 22 12x − 22 + ; A = 12, B = 2 so = =3+ 2 , − 4x + 4 x − 4x + 4 (x − 2)2 x − 2 (x − 2)2 1 1 dx + 2 dx = 3x + 12 ln |x − 2| − 2/(x − 2) + C 3dx + 12 (x − 2)2 x−2 x2

17.

us uf

A B C 1 1 1 = + + ; A = −1, B = , C = so x(x + 1)(x − 1) x x+1 x−1 2 2 1 1 1 1 1 1 1 dx + dx + dx = − ln |x| + ln |x + 1| + ln |x − 1| + C − x 2 x+1 2 x−1 2 2 1 (x + 1)(x − 1) 1 |x2 − 1| = ln + C = ln +C 2 x 2 x2 2

14.

19.

3x − 2 3x − 2 A B x2 =1+ 2 , = + ; A = −1, B = 4 so x2 − 3x + 2 x − 3x + 2 (x − 1)(x − 2) x−1 x−2 1 1 dx − dx + 4 dx = x − ln |x − 1| + 4 ln |x − 2| + C x−1 x−2 x5 + 2x2 + 1 2x2 + x + 1 2 + 1 + = x , x3 − x x3 − x

18.

A B C 2x2 + x + 1 = + + ; A = −1, B = 1, C = 2 so x(x + 1)(x − 1) x x+1 x−1 1 1 1 dx + dx + 2 dx (x2 + 1)dx − x x+1 x−1

(x + 1)(x − 1)2 1 3 1 3 +C = x + x − ln |x| + ln |x + 1| + 2 ln |x − 1| + C = x + x + ln 3 3 x

2x5 − x3 − 1 28x − 1 = 2x2 + 7 + 3 , 3 x − 4x x − 4x

20.

28x − 1 A B C 1 57 55 = + + ;A= ,B=− ,C= so x(x + 2)(x − 2) x x+2 x−2 4 8 8 1 57 1 55 1 1 dx − dx + dx (2x2 + 7) dx + 4 x 8 x+2 8 x−2 2 3 1 57 55 x + 7x + ln |x| − ln |x + 2| + ln |x − 2| + C 3 4 8 8

uh a

B C A 2x2 + 3 + = + ; A = 3, B = −1, C = 5 so x(x − 1)2 x x − 1 (x − 1)2 1 1 1 dx = 3 ln |x| − ln |x − 1| − 5/(x − 1) + C 3 dx − dx + 5 x x−1 (x − 1)2

21.

343

Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Exercise Set 8.5

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23.

Chapter 8

B C A x2 + x − 16 + + = ; A = −1, B = 2, C = −1 so (x + 1)(x − 3)2 x + 1 x − 3 (x − 3)2 1 1 1 dx + 2 dx − dx − (x − 3)2 x−3 x+1 = − ln |x + 1| + 2 ln |x − 3| +

(x − 3)2 1 1 + C = ln + +C x−3 |x + 1| x−3

A 2x2 − 2x − 1 C B ; A = 3, B = 1, C = −1 so = + 2+ x x x−1 x2 (x − 1) 1 1 1 1 3 dx − dx + dx = 3 ln |x| − − ln |x − 1| + C x x2 x−1 x

24.

3x2 − x + 1 C A B = + 2+ ; A = 0, B = −1, C = 3 so 2 x (x − 1) x x x−1 1 1 dx + 3 dx = 1/x + 3 ln |x − 1| + C − 2 x x−1

us uf

22.

Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

344

A C 2x2 + 3x + 3 B = + ; A = 2, B = −1, C = 2 so + (x + 1)3 x + 1 (x + 1)2 (x + 1)3 1 1 1 1 1 dx − dx + 2 dx = 2 ln |x + 1| + +C 2 − x+1 (x + 1)2 (x + 1)3 x + 1 (x + 1)2

1 A Bx + C = + 2 ; A = 1, B = −1, C = 0 so x(x2 + 1) x x +1 1 1 1 x2 2 + 1) + C = dx = ln |x| − ln(x ln +C x3 + x 2 2 x2 + 1

28.

A Bx + C 2x2 − 1 = + 2 ; A = −14/17, B = 12/17, C = 3/17 so 2 (4x − 1)(x + 1) 4x − 1 x +1 2x2 − 1 7 6 3 dx = − ln |4x − 1| + ln(x2 + 1) + tan−1 x + C (4x − 1)(x2 + 1) 34 17 17

27.

26.

B A C x2 + = + ; A = 1, B = −4, C = 4 so 3 2 (x + 2) x + 2 (x + 2) (x + 2)3 1 2 1 4 1 dx − 4 − dx + 4 dx = ln |x + 2| + +C x+2 (x + 2)2 (x + 2)3 x + 2 (x + 2)2

25.

x3 + 3x2 + x + 9 Ax + B Cx + D = 2 + 2 ; A = 0, B = 3, C = 1, D = 0 so (x2 + 1)(x2 + 3) x +1 x +3 3 x + 3x2 + x + 9 1 dx = 3 tan−1 x + ln(x2 + 3) + C (x2 + 1)(x2 + 3) 2

uh a

29.

30.

Ax + B Cx + D x3 + x2 + x + 2 = 2 + 2 ; A = D = 0, B = C = 1 so 2 2 (x + 1)(x + 2) x +1 x +2 3 x + x2 + x + 2 1 dx = tan−1 x + ln(x2 + 2) + C (x2 + 1)(x2 + 2) 2

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Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Exercise Set 8.5

x4 + 6x3 + 10x2 + x x = x2 + 2 , x2 + 6x + 10 x + 6x + 10 x u−3 x dx = du, dx = 2 2 x + 6x + 10 (x + 3) + 1 u2 + 1 =

us uf

32.

x3 − 3x2 + 2x − 3 x =x−3+ 2 , 2 x +1 x +1 3 1 1 x − 3x2 + 2x − 3 dx = x2 − 3x + ln(x2 + 1) + C 2 x +1 2 2

u=x+3

1 ln(u2 + 1) − 3 tan−1 u + C1 2

31.

345

x4 + 6x3 + 10x2 + x 1 1 dx = x3 + ln(x2 + 6x + 10) − 3 tan−1 (x + 3) + C 2 2 x + 6x + 10 3 1 A B 1 dx, and = + ; A = −1/6, 33. Let x = sin θ to get x2 + 4x − 5 (x + 5)(x − 1) x+5 x−1 1 1 1 1 1 1 x − 1 1 − sin θ + C = + C. B = 1/6 so we get − dx + dx = ln ln 6 x+5 6 x−1 6 x + 5 6 5 + sin θ et 1 34. Let x = et ; then dt = dx, 2t 2 e −4 x −4

35. V = π

18x2 − 81 x4 x4 =1+ 4 , dx, 4 2 2 2 (9 − x ) x − 18x + 81 x − 18x2 + 81

A B 1 = + ; A = −1/4, B = 1/4 so (x + 2)(x − 2) x+2 x−2 1 1 1 1 1 x − 2 1 et − 2 − dx + dx = ln + C = ln + C. 4 x+2 4 x−2 4 x + 2 4 et + 2

18x2 − 81 18x2 − 81 A C B D = = + ; + + 2 2 (9 − x ) (x + 3)2 (x − 3)2 x + 3 (x + 3)2 x − 3 (x − 3)2

9 9 9 9 A = − , B = , C = , D = so 4 4 4 4 2 9/4 9 19 9 9 9/4 V = π x − ln |x + 3| − =π + ln |x − 3| − − ln 5 x+3 4 x−3 0 5 4 4

ln 5

36. Let u = e to get

− ln 5

dx = 1 + ex

ln 5

− ln 5

A B 1 = + ; A = 1, B = −1; u(1 + u) u 1+u

1/5

du , u(1 + u)

5 du = (ln u − ln(1 + u)) = ln 5 u(1 + u) 1/5

Ax + B x2 + 1 Cx + D = 2 ; A = 0, B = 1, C = D = −2 so + 2 2 2 (x + 2x + 3) x + 2x + 3 (x + 2x + 3)2

uh a 37.

ex dx = ex (1 + ex )

x2 + 1 dx = 2 (x + 2x + 3)2

1 dx − (x + 1)2 + 2

2x + 2 dx (x2 + 2x + 3)2

x+1 1 = √ tan−1 √ + 1/(x2 + 2x + 3) + C 2 2

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346

38.

Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Chapter 8

x5 + x4 + 4x3 + 4x2 + 4x + 4 Ax + B Ex + F Cx + D = 2 + 2 ; + 2 2 3 2 (x + 2) x +2 (x + 2) (x + 2)3

us uf

A = B = 1, C = D = E = F = 0 so √ x+1 1 1 2 −1 √ + 2) + (x/ 2) + C tan dx = ln(x x2 + 2 2 2 39. x4 − 3x3 − 7x2 + 27x − 18 = (x − 1)(x − 2)(x − 3)(x + 3),

1 A B C D = + + + ; (x − 1)(x − 2)(x − 3)(x + 3) x−1 x−2 x−3 x+3

A = 1/8, B = −1/5, C = 1/12, D = −1/120 so 1 1 dx 1 1 ln |x − 3| − ln |x + 3| + C = ln |x − 1| − ln |x − 2| + 120 x4 − 3x3 − 7x2 + 27x − 18 8 5 12

40. 16x3 − 4x2 + 4x − 1 = (4x − 1)(4x2 + 1),

1 A Bx + C = + 2 ; A = 4/5, B = −4/5, C = −1/5 so 4x + 1 (4x − 1)(4x2 + 1) 4x − 1 1 1 1 dx tan−1 (2x) + C = ln |4x − 1| − ln(4x2 + 1) − 10 16x3 − 4x2 + 4x − 1 5 10

uh a

41. (a) x4 + 1 = (x4 + 2x2 + 1) − 2x2 = (x2 + 1)2 − 2x2 √ √ = [(x2 + 1) + 2x][(x2 + 1) − 2x] √ √ √ √ = (x2 + 2x + 1)(x2 − 2x + 1); a = 2, b = − 2 x Ax + B Cx + D √ √ √ √ (b) = + ; (x2 + 2x + 1)(x2 − 2x + 1) x2 + 2x + 1 x2 − 2x + 1 √ √ 2 2 A = 0, B = − , C = 0, D = so 4 4 √ 1 √ 1 2 1 2 x 1 1 √ √ dx + dx dx = − 4 2 2 4 0 x + 2x + 1 4 0 x − 2x + 1 0 x +1 √ 1 √ 1 2 1 2 1 √ √ =− dx + dx 2 4 0 (x − 2/2)2 + 1/2 4 0 (x + 2/2) + 1/2 √ 1+√2/2 √ 1−√2/2 2 1 2 1 =− du + du √ √ 2 2 4 u + 1/2 4 − 2/2 u + 1/2 2/2 1+√2/2 1−√2/2 √ √ 1 1 + tan−1 2u √ = − tan−1 2u √ 2 2 2/2 − 2/2 1 √ √ π 1 1 1 π + tan−1 ( 2 − 1) − − = − tan−1 ( 2 + 1) + 2 2 4 2 2 4 √ √ π 1 = − [tan−1 ( 2 + 1) − tan−1 ( 2 − 1)] 4 2 √ √ π 1 = − [tan−1 (1 + 2) + tan−1 (1 − 2)] 4 2 √ √ π 1 (1 + 2) + (1 − 2) −1 √ √ = − tan (Exercise 78, Section 7.6) 4 2 1 − (1 + 2)(1 − 2) π 1 π π π 1 = = − tan−1 1 = − 4 2 4 2 4 8

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Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Exercise Set 8.6

B 1 1 A 1 + ;A = ,B = so = 2 −x a−x a+x 2a 2a a + x 1 1 1 1 1 +C + dx = (− ln |a − x| + ln |a + x| ) + C = ln 2a a−x a+x 2a 2a a − x

EXERCISE SET 8.6

1 x +C ln 5 + 2x 5

4. Formula (66): −

1 2 2. Formula (62): + ln |2 − 3x| + C 9 2 − 3x

3 4x + ln |−1 + 4x| + C 1. Formula (60): 16 3. Formula (65):

1 − 5x 1 +C − 5 ln x x

√ 2 (−x − 4) 2 − x + C 3

1 (x + 1)(−3 + 2x)3/2 + C 5

6. Formula (105):

7. Formula (108):

√ 1 4 − 3x − 2 ln √ +C 2 4 − 3x + 2

8. Formula (108): tan−1

5. Formula (102):

√

x + √5 1 √ +C 9. Formula (69): √ ln 2 5 x − 5

3x − 4 +C 2

1 x − 3 +C 10. Formula (70): ln 6 x + 3

x 2 3 x − 3 − ln x + x2 − 3 + C 2 2 √ x2 + 5 12. Formula (93): − + ln(x + x2 + 5) + C x x 2 13. Formula (95): x + 4 − 2 ln(x + x2 + 4) + C 2 √ x x x2 − 2 9 +C 15. Formula (74): 9 − x2 + sin−1 + C 14. Formula (90): − 2 2 3 2x

11. Formula (73):

√

4 − x2 x − sin−1 + C x 2 √ 3 + √9 − x2 √ √ 17. Formula (79): 3 − x2 − 3 ln +C x

16. Formula (80): −

√

18. Formula (117): −

uh a

20. Formula (40): −

21. Formula (50):

23. Formula (42):

6x − x2 +C 3x

19. Formula (38): −

1 1 sin(5x) + sin x + C 10 2

1 1 cos(7x) + cos(3x) + C 14 6

x4 [4 ln x − 1] + C 16

us uf

42.

347

√ 1 22. Formula (50): 4 x ln x − 1 + C 2

e−2x (−2 sin(3x) − 3 cos(3x)) + C 13

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348

Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Chapter 8

ex (cos(2x) + 2 sin(2x)) + C 5 u du 4 1 1 4 − 3e2x + C = + ln 25. u = e2x , du = 2e2x dx, Formula (62): 2 (4 − 3u)2 18 4 − 3e2x

√ du 1 −1 3 x = tan +C u2 + 4 3 2

√ 2 3 27. u = 3 x, du = √ dx, Formula (68): 3 2 x

1 4

28. u = sin 4x, du = 4 cos 4xdx, Formula (68):

1 3

√

du sin 4x 1 tan−1 +C = 9 + u2 12 3

du 1 = ln 3x + 9x2 − 4 + C 3 u2 − 4

29. u = 3x, du = 3dx, Formula (76):

us uf

26. u = sin 2x, du = 2 cos 2xdx, Formula (116):

du 1 sin 2x = ln +C 2u(3 − u) 6 3 − sin 2x

24. Formula (43):

√

31. u = 3x2 , du = 6xdx, u2 du = 54x5 dx, Formula (81): u2 du 1 3x2 x2 5 √ sin−1 √ + C =− 5 − 9x4 + 36 108 54 5 5 − u2

1 du √ =− 3 − 4x2 + C 2 3x 3−u

32. u = 2x, du = 2dx, Formula (83): 2

sin2 u du =

33. u = ln x, du = dx/x, Formula (26):

1 2

34. u = e−2x , du = −2e−2x , Formula (27): − 1 4

35. u = −2x, du = −2dx, Formula (51):

1 (−2x − 1)e−2x + C 4

ln u du =

1 1 (u ln u − u) + C = (5x − 1)[ln(5x − 1) − 1] + C 5 5

uh a

1 39. u = 4x , du = 8xdx, Formula (70): 8

√

1 40. u = 2ex , du = 2ex dx, Formula (69): 2

1 1 cos2 u du = − e−2x − sin 2e−2x + C 4 8

ueu du =

37. u = cos 3x, du = −3 sin 3x, Formula (67): − 1 dx, Formula (105): x

1 1 ln x + sin(2 ln x) + C 2 4

1 36. u = 5x − 1, du = 5dx, Formula (50): 5

38. u = ln x, du =

√ 2x2 , du = 2 2xdx, Formula (72): √ 1 x2 4 3 √ u2 + 3 du = 2x + 3 + √ ln 2x2 + 2x4 + 3 + C 4 2 2 4 2

30. u =

cos 3x 1 du 1 +C =− + ln u(u + 1)2 3 1 + cos 3x 1 + cos 3x

√ u du 1 (2 ln x + 1) 4 ln x − 1 + C = 12 4u − 1

2 4x − 1 du 1 +C ln = u2 − 1 16 4x2 + 1 2ex + √3 du 1 √ +C = √ ln x 3 − u2 4 3 2e − 3

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us uf

41. u = 2ex , du = 2ex dx, Formula (74): √ √ 1 1 3 1 3 3 − u2 du = u 3 − u2 + sin−1 (u/ 3) + C = ex 3 − 4e2x + sin−1 (2ex / 3) + C 2 4 4 2 4

44. u =

√

42. u = 3x, du = 3dx, Formula (80): √ √ √ 4 − 9x2 4 − u2 du 4 − u2 −1 3 − 3 sin (u/2) + C = − = −3 − 3 sin−1 (3x/2) + C 2 u x u 43. u = 3x, du = 3dx, Formula (112):

1 5 u−5 5 5 1 25 −1 2 2 +C u− u − u du = u−u + sin 5 3 3 6 6 3 216 18x − 5 18x − 5 25 −1 2 = +C sin 5x − 9x + 5 36 216

5 dx, Formula (117): √ √ (u/ 5) − u2 du x − 5x2 √ √ +C =− + C = −2 x u/(2 5) u (u/ 5) − u2 5x, du =

45. u = 3x, du = 3dx, Formula (44): 1 1 1 u sin u du = (sin u − u cos u) + C = (sin 3x − 3x cos 3x) + C 9 9 9 √ √ √ √ 46. u = x, u2 = x, 2udu = dx, Formula (45): 2 u cos u du = 2 cos x + 2 x sin x + C

√ √ ueu du = −2( x + 1)e− x + C

√ 47. u = − x, u2 = x, 2udu = dx, Formula (51): 2

349

Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Exercise Set 8.6

48. u = 2 − 3x2 , du = −6xdx, Formula (50): 1 1 1 1 ln u du = − (u ln u − u) + C = − ((2 − 3x2 ) ln(2 − 3x2 ) + (2 − 3x2 ) + C − 6 6 6 6

49. x2 + 4x − 5 = (x + 2)2 − 9; u = x + 2, du = dx, Formula (70): 1 x − 1 du 1 u − 3 + C = ln +C = ln u2 − 9 6 u + 3 6 x + 5

uh a

50. x2 + 2x − 3 = (x + 1)2 − 4, u = x + 1, du = dx, Formula (77): 1 4 − u2 du = u 4 − u2 + 2 sin−1 (u/2) + C 2 1 = (x + 1) 3 − 2x − x2 + 2 sin−1 ((x + 1)/2) + C 2

51. x2 − 4x − 5 = (x − 2)2 − 9, u = x − 2, du = dx, Formula (77): u+2 u u du du √ √ du = +2 √ = − 9 − u2 + 2 sin−1 + C 2 2 2 3 9−u 9−u 9−u x−2 = − 5 + 4x − x2 + 2 sin−1 +C 3

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350

Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Chapter 8

us uf

52. x2 + 6x + 13 = (x + 3)2 + 4, u = x + 3, du = dx, Formula (71): 1 (u − 3) du 3 1 3 = ln(u2 + 4) − tan−1 (u/2) + C = ln(x2 + 6x + 13) − tan−1 ((x + 3)/2) + C 2 2 2 2 u2 + 4 √ 53. u = x − 2, x = u2 + 2, dx = 2u du; 2 4 2 4 2 2 2u (u + 2)du = 2 (u4 + 2u2 )du = u5 + u3 + C = (x − 2)5/2 + (x − 2)3/2 + C 5 3 5 3

√ 54. u = x + 1, x = u2 − 1, dx = 2u du; √ 2 2 2 (u2 − 1)du = u3 − 2u + C = (x + 1)3/2 − 2 x + 1 + C 3 3

√ 55. u = x3 + 1, x3 = u2 − 1, 3x2 dx = 2u du; 2 2 5 2 3 2 3 2 2 u2 (u2 − 1)du = (u4 − u2 )du = u − u +C = (x + 1)5/2 − (x3 + 1)3/2 + C 3 3 15 9 15 9

√ 56. u = x3 − 1, x3 = u2 + 1, 3x2 dx = 2u du; 2 1 2 2 du = tan−1 u + C = tan−1 x3 − 1 + C 2 u +1 3 3 3

57. u = x1/6 , x = u6 , dx = 6u5 du; 6u5 u3 1 2 du = 6 u −u+1− du du = 6 u3 + u2 u+1 u+1

= 2x1/2 − 3x1/3 + 6x1/6 − 6 ln(x1/6 + 1) + C 58. u = x

5u4 du = 5 u5 − u3

, x = u , dx = 5u du;

1/5

59. u = x1/4 , x = u4 , dx = 4u3 du; 4

60. u = x1/3 , x = u3 , dx = 3u2 du; 3

1 du = 4 u(1 − u) u4 du = 3 3 u +1

u 5 du = ln |x2/5 − 1| + C u2 − 1 2

1 1 x1/4 + du = 4 ln +C u 1−u |1 − x1/4 |

u−

u 3 u +1

du,

uh a

u u −1/3 (1/3)u + 1/3 = = + 2 so u3 + 1 (u + 1)(u2 − u + 1) u+1 u −u+1 1 u+1 u du = 3u + − 2 du 3 u− 3 u +1 u+1 u −u+1 √ 3 2 2u − 1 1 u + ln |u + 1| − ln(u2 − u + 1) − 3 tan−1 √ +C 2 2 3 √ 2x1/3 − 1 1 3 √ = x2/3 + ln |x1/3 + 1| − ln(x2/3 − x1/3 + 1) − 3 tan−1 +C 2 2 3 =

61. u = x1/6 , x = u6 , dx = 6u5 du; 1 u3 du = 2x1/2 + 3x1/3 + 6x1/6 + 6 ln |x1/6 − 1| + C 6 du = 6 u2 + u + 1 + u−1 u−1

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Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Exercise Set 8.6

us uf

√ 62. u = x, x = u2 , dx = 2u du; 2 √ √ u +u 2 du = −x − 4 x − 4 ln | x − 1| + C −2 du = −2 u+2+ u−1 u−1 √ 63. u = 1 + x2 , x2 = u2 − 1, 2x dx = 2u du, x dx = u du; 1 (u2 − 1)du = (1 + x2 )3/2 − (1 + x2 )1/2 + C 3

67.

2u 1 − u2 1+ + 2 1+u 1 + u2

68.

1 2u 2+ 1 + u2

2 du = 1 + u2

1 du = ln | tan(x/2) + 1| + C u+1

1 du u2 + u + 1

1 2 du = √ tan−1 (u + 1/2)2 + 3/4 3

2 tan(x/2) + 1 √ 3

√ √ √ ueu du = 2 xe x − 2e x + C

x, x = u2 , dx = 2u du, Formula (51): 2

√ √ √ u sin u du = 2 sin x − 2 x cos x + C

√

x, x = u2 , dx = 2u du, Formula (44): 2

66. u =

√

64. u = (x + 3)1/5 , x = u5 − 3, dx = 5u4 du; 5 15 5 (u8 − 3u3 )du = (x + 3)9/5 − (x + 3)4/5 + C 9 4 65. u =

dθ = 1 − cos θ

69. u = tan(θ/2),

1 1 du = − + C = − cot(θ/2) + C u2 u

70. u = tan(x/2), 2 1 1 2 2 du = du = dz (z = u + 4/3) 3u2 + 8u − 3 3 (u + 4/3)2 − 25/9 3 z 2 − 25/9 1 tan(x/2) − 1/3 1 z − 5/3 + C = ln +C = ln 5 z + 5/3 5 tan(x/2) + 3

71. u = tan(x/2), 2

1 − u2 du; (3u2 + 1)(u2 + 1)

uh a

(0)u + 2 (0)u − 1 2 1 1 − u2 = + 2 = 2 − 2 so 2 2 2 3u + 1 u +1 3u + 1 u + 1 (3u + 1)(u + 1) √

4 cos x 3 tan(x/2) − x + C dx = √ tan−1 2 − cos x 3

1 72. u = tan(x/2), 2

1 − u2 1 du = u 2

(1/u − u)du =

351

1 1 ln | tan(x/2)| − tan2 (x/2) + C 2 4

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Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

73. 2

x 1 1 t (Formula (65), a = 4, b = −1) dt = ln t(4 − t) 4 4−t 2 x 1 1 x 1 x x ln = − ln 1 = ln , ln = 0.5, ln = 2, 4 4−x 4 4−x 4 4−x 4−x

Chapter 8

us uf

352

x = e2 , x = 4e2 − e2 x, x(1 + e2 ) = 4e2 , x = 4e2 /(1 + e2 ) ≈ 3.523188312 4−x x x √ 1 √ (Formula (108), a = −1, b = 2) 74. dt = 2 tan−1 2t − 1 1 t 2t − 1 1 √ √ = 2 tan−1 2x − 1 − tan−1 1 = 2 tan−1 2x − 1 − π/4 , √ √ √ 2(tan−1 2x − 1 − π/4) = 1, tan−1 2x − 1 = 1/2 + π/4, 2x − 1 = tan(1/2 + π/4),

75. A = 0

4 1 25 −1 x 2 2 25 − x dx = x 25 − x + sin 2 2 5 0 =6+

4 25 sin−1 ≈ 17.59119023 2 5

9x2 − 4 dx; u = 3x,

76. A = 2/3

(Formula (74), a = 5)

x = [1 + tan2 (1/2 + π/4)]/2 ≈ 6.307993516

6 1 1 2 2 u u − 4 − 2 ln u + u − 4 − 4 du = (Formula (73), a2 = 4) 3 2 2 2 √ √ √ 2 1 √ 3 32 − 2 ln(6 + 32) + 2 ln 2 = 4 2 − ln(3 + 2 2) ≈ 4.481689467 = 3 3

1 A= 3

1 dx; u = 4x, 25 − 16x2 0 u + 5 4 1 4 1 1 1 A= ln ln 9 ≈ 0.054930614 (Formula (69), a = 5) du = = 4 0 25 − u2 40 u − 5 0 40

77. A =

√

x ln x dx =

79. V = 2π

π/2

80. V = 2π

(Formula (50), n = 1/2) 1

(Formula (45))

√ 4π (3x + 8)(x − 4)3/2 x x − 4 dx = 15 =

8 (Formula (102), a = −4, b = 1) 4

1024 π ≈ 214.4660585 15

uh a

3 ln x − 1 2

π/2 x cos x dx = 2π(cos x + x sin x) = π(π − 2) ≈ 3.586419094

4 (12 ln 4 − 7) ≈ 4.282458815 9

4 3/2 x 9

78. A =

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Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Exercise Set 8.6

353

xe−x dx; u = −x,

81. V = 2π

V = 2π

−3 ue du = 2πe (u − 1) = 2π(1 − 4e−3 ) ≈ 5.031899801 u

5 π 2 x ln x dx = x (2 ln x − 1) 2 1

82. V = 2π 1

= π(25 ln 5 − 12) ≈ 88.70584621

(Formula (50), n = 1)

1 + 16x2 dx; u = 4x,

83. L =

(Formula (51))

us uf

−3

0 8

1 + u2 du =

84. L =

1 4 √

8 u 1 1 + u2 + ln u + 1 + u2 2 2 0 √ 1 65 + ln(8 + 65) ≈ 8.409316783 8

3 3 + √x2 + 9 x2 + 9 − 3 ln x 1 1 √ √ √ 3 + 10 √ ≈ 3.891581644 (Formula (89), a = 3) = 3 2 − 10 + 3 ln 1+ 2

1 + 9/x2 dx =

85. S = 2π

√

x2 + 9 dx = x

(sin x) 1 + cos2 x dx; u = cos x,

0 −1

1 + u2 du = 4π

86. S = 2π 1 16

2 + ln(1 +

1 u 1 1 + u2 + ln u + 1 + u2 a2 = 1 2 2 0

√

2) ≈ 14.42359945

(Formula (72))

4√ 4 1 x +1 4 1 + 1/x dx = 2π dx; u = x2 , x x3 1 √ √ 16 u2 + 1 u2 + 1 2 du = π − + ln u + u + 1 u2 u 1 √ √ √ 257 16 + 257 √ =π 2− ≈ 9.417237485 (Formula (93), a2 = 1) + ln 16 1+ 2

S=π

√

1 + u2 du = 4π

= 2π

S = −2π

(Formula (72), a2 = 1)

1 4

40 166 20 sin2 t cos7 t − cos7 t + 9 63 63

(b)

s(t) 4 3 2 1 t 3

uh a

=−

20 cos6 u sin3 u du

87. (a) s(t) = 2 +

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Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Chapter 8

a(u) du = â&#x2C6;&#x2019; 0

1 1 â&#x2C6;&#x2019;t 1 1 3 1 e cos 2t + eâ&#x2C6;&#x2019;t sin 2t + eâ&#x2C6;&#x2019;t cos 6t â&#x2C6;&#x2019; eâ&#x2C6;&#x2019;t sin 6t + â&#x2C6;&#x2019; 10 5 74 37 10 74

s(t) = 10 +

88. (a) v(t) =

v(u) du

us uf

354

=â&#x2C6;&#x2019; s(t)

t 2

89. (a)

1 + tan(x/2) 1 + u 2 1 +C dx = + C = ln du = ln sec x dx = cos x 1 â&#x2C6;&#x2019; u2 1 â&#x2C6;&#x2019; u 1 â&#x2C6;&#x2019; tan(x/2) cos(x/2) + sin(x/2) cos(x/2) + sin(x/2) + C = ln 1 + sin x + C = ln cos(x/2) â&#x2C6;&#x2019; sin(x/2) cos(x/2) + sin(x/2) cos x

12 10 8 6 4 2

(b)

2 35 â&#x2C6;&#x2019;t 6 â&#x2C6;&#x2019;t 16 3 â&#x2C6;&#x2019;t 343866 e cos 2t â&#x2C6;&#x2019; eâ&#x2C6;&#x2019;t sin 2t + e cos 6t + e sin 6t + t+ 50 25 2738 1369 185 34225

= ln |sec x + tan x| + C

Ď&#x20AC; x x tan + tan 1 + tan x 4 2 2 (b) tan + = x = Ď&#x20AC; x 4 2 1 â&#x2C6;&#x2019; tan tan 1 â&#x2C6;&#x2019; tan 2 4 2 1 dx = 1/u du = ln | tan(x/2)| + C but 90. csc x dx = sin x

1 sin2 (x/2) 1 (1 â&#x2C6;&#x2019; cos x)/2 1 1 â&#x2C6;&#x2019; cos x = ln ln = ln ; also, 2 cos2 (x/2) 2 (1 + cos x)/2 2 1 + cos x

ln | tan(x/2)| =

Ď&#x20AC;

1 â&#x2C6;&#x2019; cos2 x 1 1 1 â&#x2C6;&#x2019; cos x 1 â&#x2C6;&#x2019; cos x = â&#x2C6;&#x2019; ln | csc x + cot x| = = so ln 2 2 (1 + cos x) (csc x + cot x) 2 1 + cos x 1 + cos x

â&#x2C6;&#x161; 91. Let u = tanh(x/2) then cosh(x/2) = 1/ sech(x/2) = 1/ 1 â&#x2C6;&#x2019; tanh2 (x/2) = 1/ 1 â&#x2C6;&#x2019; u2 , â&#x2C6;&#x161; sinh(x/2) = tanh(x/2) cosh(x/2) = u/ 1 â&#x2C6;&#x2019; u2 , so sinh x = 2 sinh(x/2) cosh(x/2) = 2u/(1 â&#x2C6;&#x2019; u2 ),

cosh x = cosh2 (x/2) + sinh2 (x/2) = (1 + u2 )/(1 â&#x2C6;&#x2019; u2 ), x = 2 tanhâ&#x2C6;&#x2019;1 u, dx = [2/(1 â&#x2C6;&#x2019; u2 )]du; dx 2u + 1 2 tanh(x/2) + 1 2 1 2 â&#x2C6;&#x161; + C = â&#x2C6;&#x161; tanâ&#x2C6;&#x2019;1 + C. = du = â&#x2C6;&#x161; tanâ&#x2C6;&#x2019;1 â&#x2C6;&#x161; 2 cosh x + sinh x u2 + u + 1 3 3 3 3

uh a

EXERCISE SET 8.7

1. exact value = 14/3 â&#x2030;&#x2C6; 4.666666667 (a) 4.667600663, |EM | â&#x2030;&#x2C6; 0.000933996 (b) 4.664795679, |ET | â&#x2030;&#x2C6; 0.001870988 (c) 4.666651630, |ES | â&#x2030;&#x2C6; 0.000015037

2. exact value = 2 (a) 1.998377048, |EM | â&#x2030;&#x2C6; 0.001622952 (b) 2.003260982, |ET | â&#x2030;&#x2C6; 0.003260982 (c) 2.000072698, |ES | â&#x2030;&#x2C6; 0.000072698

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Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Exercise Set 8.7

(a) 2.008248408, |EM | ≈ 0.008248408 (b) 1.983523538, |ET | ≈ 0.016476462 (c) 2.000109517, |ES | ≈ 0.000109517

1 ln 5 ≈ 0.804718956 2 (a) 0.801605339, |EM | ≈ 0.003113617 (b) 0.811019505, |ET | ≈ 0.006300549 (c) 0.805041497, |ES | ≈ 0.000322541

5. exact value = e−1 − e−3 ≈ 0.318092373

6. exact value =

(a) 0.317562837, |EM | ≈ 0.000529536 (b) 0.319151975, |ET | ≈ 0.001059602 (c) 0.318095187, |ES | ≈ 0.000002814

1 15 x + 1, f (x) = − (x + 1)−3/2 , f (4) (x) = − (x + 1)−7/2 ; K2 = 1/4, K4 = 15/16 16 4

27 (1/4) = 0.002812500 2400 243 (c) |ES | ≤ (15/16) ≈ 0.000126563 180 × 104

(a) |EM | ≤

(b) |ET | ≤

27 (1/4) = 0.005625000 1200

√

us uf

(a) 0.841821700, |EM | ≈ 0.000350715 (b) 0.840769642, |ET | ≈ 0.000701343 (c) 0.841471453, |ES | ≈ 0.000000468

27 (3/4) = 0.008437500 2400 243 (c) |ES | ≤ (105/16) ≈ 0.000885938 180 × 104

√ 3 105 −9/2 ; K2 = 3/4, K4 = 105/16 8. f (x) = 1/ x, f (x) = x−5/2 , f (4) (x) = x 4 16 (b) |ET | ≤

(a) |EM | ≤

27 (3/4) = 0.016875000 1200

9. f (x) = sin x, f (x) = − sin x, f (4) (x) = sin x; K2 = K4 = 1

π3 (1) ≈ 0.012919282 2400 π5 (c) |ES | ≤ (1) ≈ 0.000170011 180 × 104

(a) |EM | ≤

(b) |ET | ≤

π3 (1) ≈ 0.025838564 1200

10. f (x) = cos x, f (x) = − cos x, f (4) (x) = cos x; K2 = K4 = 1 1 (1) ≈ 0.000416667 2400 1 (1) ≈ 0.000000556 (c) |ES | ≤ 180 × 104

(b) |ET | ≤

1 (1) ≈ 0.000833333 1200

(a) |EM | ≤

11. f (x) = e−x , f (x) = f (4) (x) = e−x ; K2 = K4 = e−1 8 (e−1 ) ≈ 0.001226265 2400 32 (e−1 ) ≈ 0.000006540 (c) |ES | ≤ 180 × 104

(b) |ET | ≤

8 (e−1 ) ≈ 0.002452530 1200

(a) |EM | ≤

uh a

12. f (x) = 1/(2x + 3), f (x) = 8(2x + 3)−3 , f (4) (x) = 384(2x + 3)−5 ; K2 = 8, K4 = 384 8 (8) ≈ 0.026666667 2400 32 (c) |ES | ≤ (384) ≈ 0.006826667 180 × 104

(a) |EM | ≤

(b) |ET | ≤

4. exact value = sin(1) ≈ 0.841470985

3. exact value = 2

7. f (x) =

355

8 (8) ≈ 0.053333333 1200

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Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Chapter 8

(8)(e−1 ) (24)(10−6 )

≈ 33.5; n = 34

1/2

≈ 58.1; n = 59

1/2

≈ 50.8; n = 51

≈ 6.4; n = 8 (b) n >

1/4 ≈ 1.5; n = 2

1/2

(32)(e−1 ) (c) n > (180)(10−6 )

(π 3 )(1) (12)(10−3 )

1/4

≈ 6.5; n = 7

(1)(1) (c) n > (180)(10−3 ) 17. (a) n >

(b) n >

1/2

≈ 350.2; n = 351

1/4

(1)(1) (12)(10−3 )

1/2 ≈ 9.1; n = 10

(1)(1) (24)(10−3 )

≈ 35.9; n = 36

(π 5 )(1) (c) n > (180)(10−3 )

(27)(3/4) (b) n > (12)(5 × 10−4 )

1/2

16. (a) n >

(b) n >

(π 3 )(1) (24)(10−3 )

(8)(e−1 ) (12)(10−6 )

1/2 ≈ 495.2; n = 496

≈ 15.99; n = 16

15. (a) n >

(27)(1/4) (b) n > (12)(5 × 10−4 )

us uf

1/2 (27)(3/4) 14. (a) n > ≈ 41.1; n = 42 (24)(5 × 10−4 ) 1/4 (243)(105/16) (c) n > ≈ 11.5; n = 12 (180)(5 × 10−4 )

1/2 (27)(1/4) ≈ 23.7; n = 24 13. (a) n > (24)(5 × 10−4 ) 1/4 (243)(15/16) (c) n > ≈ 7.1; n = 8 (180)(5 × 10−4 )

356

1/2 (8)(8) 18. (a) n > ≈ 1632.99; n = 1633 (24)(10−6 ) 1/4 (32)(384) (c) n > ≈ 90.9; n = 92 (180)(10−6 )

(8)(8) (b) n > (12)(10−6 )

1/2 ≈ 2309.4; n = 2310

19. g(X0 ) = aX02 + bX0 + c = 4a + 2b + c = f (X0 ) = 1/X0 = 1/2; similarly 9a + 3b + c = 1/3, 16a + 4b + c = 1/4. Three equations in three unknowns, with solution a = 1/24, b = −3/8, c = 13/12, g(x) = x2 /24 − 3x/8 + 13/12. 4 2 25 x 3x 13 dx = g(x) dx = − + 24 8 12 36 0 25 1 1 4 1 ∆x [f (X0 ) + 4f (X1 ) + f (X2 )] = + + = 3 3 2 3 4 36 20. f (X0 ) = 1 = g(X0 ) = c, f (X1 ) = 3/4 = g(X1 ) = a/36 + b/6 + c,

uh a

f (X2 ) = 1/4 = g(X2 ) = a/9 + b/3 + c, with solution a = −9/2, b = −3/4, c = 1, g(x) = −9x2 /2 − 3x/4 + 1, 1/3 g(x) dx = 17/72

1 ∆x [f (X0 ) + 4f (X1 ) + f (X2 ) = [1 + 3 + 1/4] = 17/72 3 18

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22. 1.137631378, 1.137630147

23. 2.129861595, 2.129861293

24. 2.418388347, 2.418399152

25. 0.805376152, 0.804776489

26. 1.536963087, 1.544294774

27. (a) 3.142425985, |EM | ≈ 0.000833331 (b) 3.139925989, |ET | ≈ 0.001666665 (c) 3.141592614, |ES | ≈ 0.000000040

us uf

21. 0.746824948, 0.746824133

357

Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Exercise Set 8.7

28. (a) 3.152411433, |EM | ≈ 0.010818779 (b) 3.104518326, |ET | ≈ 0.037074328 (c) 3.127008159, |ES | ≈ 0.014584495

29. S14 = 0.693147984, |ES | ≈ 0.000000803 = 8.03 × 10−7 ; the method used in Example 6 results in a value of n which ensures that the magnitude of the error will be less than 10−6 , this is not necessarily the smallest value of n. 2

30. (a) greater, because the graph of e−x is concave up on the interval (1, 2) 2

(b) less, because the graph of e−x is concave down on the interval (0, 0.5) 31. f (x) = x sin x, f (x) = 2 cos x − x sin x, |f (x)| ≤ 2| cos x| + |x| | sin x| ≤ 2 + 2 = 4 so K2 ≤ 4, (8)(4) n> (24)(10−4 )

1/2

≈ 115.5; n = 116 (a smaller n might suﬃce)

33. f (x) =

√

(1)(2e) (24)(10−4 )

x, f (x) = −

1/2

≈ 47.6; n = 48 (a smaller n might suﬃce)

K2 ≤ 2e, n >

32. f (x) = ecos x , f (x) = (sin2 x)ecos x − (cos x)ecos x , |f (x)| ≤ ecos x (sin2 x + | cos x|) ≤ 2e so

1 , lim |f (x)| = +∞ 4x3/2 x→0+

√ √ x sin x + cos x 34. f (x) = sin x, f (x) = − , lim |f (x)| = +∞ x→0+ 4x3/2 π

t (s)

v (mi/hr)

v (ft/s) 20

v dt ≈

uh a

38.

1 + 1/x4 dx ≈ 2.146822803

58.67

107.07

123.2

20 [0 + 4(58.67) + 2(88) + 4(107.07) + 123.2] ≈ 1604 ft (3)(4)

36. L =

37.

1 + cos2 x dx ≈ 3.820187623

35. L =

√

0.02

0.08

0.20

0.40

0.60

0.70

0.60

8 [0 + 4(0.02) + 2(0.08) + 4(0.20) + 2(0.40) + 4(0.60) + 2(0.70) + 4(0.60) + 0] (3)(8) ≈ 2.7 cm/s

a dt ≈

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Chapter 8

180

v dt ≈ 0

1800

40. 0

180 [0.00 + 4(0.03) + 2(0.08) + 4(0.16) + 2(0.27) + 4(0.42) + 0.65] = 37.9 mi (3)(6)

1 4 2 4 2 4 1 1800 + + + + + + ≈ 0.71 s (1/v)dx ≈ (3)(6) 3100 2908 2725 2549 2379 2216 2059

πr2 dy = π

41. V =

r2 dy ≈ π 0

16 [(8.5)2 + 4(11.5)2 + 2(13.8)2 + 4(15.4)2 + (16.8)2 ] (3)(4)

≈ 9270 cm3 ≈ 9.3 L 600

600 [0 + 4(7) + 2(16) + 4(24) + 2(25) + 4(16) + 0] = 9000 ft2 , (3)(6) 0 V = 75A ≈ 75(9000) = 675, 000 ft3 b

f (x) dx ≈ A1 + A2 + · · · + An =

43. a

b−a 1 1 1 (y0 + y1 ) + (y1 + y2 ) + · · · + (yn−1 + yn ) n 2 2 2

h dx ≈

b−a [y0 + 2y1 + 2y2 + · · · + 2yn−1 + yn ] 2n

44. right endpoint, trapezoidal, midpoint, left endpoint

42. A =

45. (a) The maximum value of |f (x)| is approximately 3.844880. (b) n = 18 (c) 0.904741

46. (a) The maximum value of |f (x)| is approximately 1.467890. (b) n = 12 (c) 1.112062 47. (a) The maximum value of |f (4) (x)| is approximately 42.551816. (b) n = 8 (c) 0.904524

48. (a) The maximum value of |f (4) (x)| is approximately 7.022710. (b) n = 8 (c) 1.111443

EXERCISE SET 8.8

improper; infinite discontinuity at x = 3 continuous integrand, not improper improper; infinite discontinuity at x = 0 improper; infinite interval of integration improper; infinite interval of integration and infinite discontinuity at x = 1 continuous integrand, not improper

uh a

1. (a) (b) (c) (d) (e) (f )

39.

us uf

358

2. (a) improper if p > 0 (b) improper if 1 ≤ p ≤ 2 (c) integrand is continuous for all p, not improper

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Exercise Set 8.8

10.

11.

12.

→+∞

1 − 2 −e + 1 = 1/2 →+∞ 2

= lim

→+∞

−

1 1 1 = + 2 2 ln2 + 2

→+∞

lim −

→−∞

1 4(2x − 1)2

1 x tan−1 →−∞ 2 2

= lim

lim

1 3x e →−∞ 3

0 →−∞

1 1 π + − tan−1 = [π/4 − (−π/2)] = 3π/8 →−∞ 2 4 2 2

= lim

→−∞

1 [−1 + 1/(2+ − 1)2 ] = −1/4 4

= lim

lim

us uf

3 5 = ln 5 3

= lim

1 2 ln2 x

→+∞

1 1 1 3 = − e 3 3 3

0 1 1 1 lim − ln(3 − 2ex ) = lim ln(3 − 2e ) = ln 3 →−∞ →−∞ 2 2 2

+∞

x3 dx converges if −∞

+∞

∞

x2 + 2

√

−∞

dx = lim

uh a

√

14.

1 4 x →+∞ 4

→+∞

x3 dx both converge; it diverges if either (or both) 0

x3 dx = lim

+∞

x3 dx and

−∞

diverges.

1 4 + = +∞ so →+∞ 4

+∞

x3 dx is divergent.

= lim 0

−∞

√ x2 + 2 = lim ( +2 + 2 − 2) = +∞ 0

→+∞

dx is divergent.

x 1 1 1 dx = lim − = lim [−1/(+2 + 3) + 1/3] = , 2 2 2 →+∞ (x + 3) 2(x + 3) 0 →+∞ 2 6 0 0 ∞ x x similarly dx = −1/6 so dx = 1/6 + (−1/6) = 0 2 + 3)2 2 + 3)2 (x (x −∞ −∞

= − ln

+−1 3 − ln ++1 5

√ √ √ lim 2 ln x = lim (2 ln + − 2 ln 2) = +∞, divergent

13.

15.

→+∞

2 1 lim − e−x →+∞ 2

lim −

→+∞

= lim

x−1 x+1

lim ln

1 1 2 lim ln(1 + x ) = lim [ln(1 + +2 ) − ln 2] = +∞, divergent →+∞ 2 →+∞ 2 −1

→+∞

lim (−e

→+∞

) = lim (−e− + 1) = 1

−x

359

+∞

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Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

360

Chapter 8

e−t π π −1 −t −1 − = , − tan dt = lim − tan (e ) = lim (e ) + →+∞ →+∞ 1 + e−2t 4 4 0 0 π

π e−t −1 −t −1 − − dt = lim − tan (e ) = lim (e ) = , + tan →−∞ →−∞ 1 + e−2t 4 4

−∞

e−t π π π dt = + = −2t 1+e 4 4 2

+∞

−∞

23.

24.

→π/2

lim − ln(cos +) = +∞, divergent

→π/2−

√ √ lim −2 9 − x = lim 2(− 9 − + + 3) = 6 →9−

0 −1

lim sin

→1−

→9−

x = lim sin−1 + = π/2 0

→1−

1 √ √ 2 = lim + (− 8 + 9 − +2 ) = − 8 lim + − 9 − x

→−3

√ lim − − 1 − 2 sin x =

→π/6

√ lim (− 1 − 2 sin + + 1) = 1

→π/6−

lim − ln(1 − tan x) = −

→π/4

25. 0

26. 0

27. 0

−1

lim − ln(1 − tan +) = +∞, divergent

→π/4−

dx = lim ln |x − 2| = lim (ln |+ − 2| − ln 2) = −∞, divergent x−2 →2− →2− 0 2 2 dx dx = lim+ −1/x = lim+ (−1/2 + 1/+) = +∞ so is divergent 2 x2 →0 →0 −2 x 3 x−1/3 dx = lim+ x2/3 →0 2

3 x−1/3 dx = lim− x2/3 →0 2 8

3 = lim+ (4 − +2/3 ) = 6, →0 2

−1

3 = lim− (+2/3 − 1) = −3/2 →0 2

x−1/3 dx = 6 + (−3/2) = 9/2

uh a

22.

lim − − ln(cos x) =

21.

3 = lim+ (4 − +2/3 ) = 6 →0 2

20.

→0+

→3+

1 = +∞, divergent +−3

19.

3 2/3 x 2

lim

= lim −1 +

18.

17.

1 lim+ − x − 3 →3

us uf

16.

+∞

−1

√ dx 3 1/3 1/3 1/3 28. = lim 3(x − 2) = lim 3[(+ − 2) − (−2) ] = 3 2, 2/3 − − →2 →2 (x − 2) 0 0 4 4 4 √ √ dx dx 3 3 1/3 similarly = lim 3(x − 2) = 3 2 so = 6 2 2/3 2/3 + →2 (x − 2) (x − 2) 2 0 2

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Exercise Set 8.8

+∞ 1 1 dx + dx where a > 0; take a = 1 for convenience, 2 2 x x 0 a 0 1 +∞ 1 1 dx = lim (−1/x) = lim (1/+ − 1) = +∞ so dx is divergent. 2 + + x x2 →0 →0 0 +∞

1 dx = x2

+∞

dx √ = x x2 − 1

30. Deﬁne 1

dx √ + x x2 − 1

+∞

dx √ where a > 1, x x2 − 1

take a = 2 for convenience to get 2 2 dx −1 √ = lim+ sec x = lim+ (π/3 − sec−1 +) = π/3, 2 →1 →1 1 x x −1 +∞ +∞ dx dx √ √ = π/2. = lim sec−1 x = π/2 − π/3 so 2 →+∞ x x2 − 1 x x −1 1 2 2

√ 0

33. 0

+∞

√

34. 0

→+∞

+∞

−3x

37. (a) 2.726585

dy dx

√ 1

du = 1 − u2

uh a

= lim tan−1 →+∞

= lim 2(1 −

√

→0+

√

+ π = 2 2

+) = 2

du = lim sin−1 u 2 →1 1−u

1 dx = lim − (3x + 1)e−3x →+∞ 9

= lim sin−1 + = →1

= 1/3 0

(b) 2.804364

4 − x2/3 4 =1+ = 2/3 ; the arc length is 2/3 x x

2 x1/3

16x2 9 + 12x2 =1+ = ; the arc length is 2 9 − 4x 9 − 4x2

ln x dx = lim+ →0

but lim + ln + = lim + + →0

π 2

dy dx

√ du √ = lim 2 u + u →0

1 − e− = 2

3/2

(d) 0.504067 8 = 12

2/3

dx = 3x

9 + 12x2 dx ≈ 3.633168 9 − 4x2

ln x dx = x ln x − x + C,

41.

→+∞

40. 1 +

= 2 lim

1 −x e (sin x − cos x) = 1/2 →+∞ 2 0

du 1 u = 2 lim tan−1 →+∞ 2 u2 + 4 2

e−x cos x dx = lim

36. A =

39. 1 +

+∞

e−x dx = − 1 − e−2x

lim

−e

dx =2 x(x + 4)

35.

→+∞

e−x √ dx = 1 − e−x

+∞

du = 2 lim

−u

+∞

32.

dx = 2

−u

+∞

31.

√ x

e− √

+∞

→0

1 ln x dx = lim+ (x ln x − x) = lim+ (−1 − + ln + + +), →0

ln + = lim (−+) = 0 so 1/+ →0+

us uf

29. Deﬁne

361

→0

ln x dx = −1 0

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Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

362

Chapter 8

+∞

43.

−3x

dx = lim

→+∞

+∞

−3x

dx = lim

→+∞

8 x−2 dx = lim 2 ln →+∞ x2 − 4 x+2

44. A = 3

+∞

45. (a) V = π

e−2x dx = −

+∞

e−x

π 2

1 − e−3x 3

= lim

→+∞

1 1 − e−3 + 3 9

1 3

us uf

ln x 1 ln x − + C, dx = − 2 x x x +∞ ln x 1 ln + 1 ln x ln x − − dx = lim dx = lim = lim − − + 1 , →+∞ 1 x2 →+∞ x2 x x 1 →+∞ + + 1 +∞ ln + 1 ln x = lim = 0 so =1 but lim →+∞ + →+∞ + x2 1

+−2 1 = 2 ln 5 = lim 2 ln − ln →+∞ 5 ++2

lim e−2x

→+∞

= π/2

42.

1 + e−2x dx, let u = e−x to get 0 1 √ √

u 1 S = −2π 1 + u2 du = 2π 1 + u2 + ln u + 1 + u2 = π 2 + ln(1 + 2) 2 2 1 0

(b) S = 2π

47. (a) For x ≥ 1, x2 ≥ x, e−x ≤ e−x +∞

−x (b) e dx = lim e−x dx = lim −e−x = lim (e−1 − e− ) = 1/e →+∞ 1 →+∞ →+∞ 1 1 +∞ 2 (c) By Parts (a) and (b) and Exercise 46(b), e−x dx is convergent and is ≤ 1/e. 1

ex 1 ≤ 48. (a) If x ≥ 0 then ex ≥ 1, 2x + 1 2x + 1 dx 1 (b) lim = lim ln(2x + 1) = +∞ →+∞ 0 2x + 1 →+∞ 2 0 +∞ (c) By Parts (a) and (b) and Exercise 46(a),

A = lim

(π/x2 ) dx = lim −(π/x) = lim (π − π/+) = π →+∞

→+∞

49. V = lim

→+∞

2π(1/x) 1 + 1/x4 dx; use Exercise 46(a) with f (x) = 2π/x, g(x) = (2π/x) 1 + 1/x4

→+∞

ex dx is divergent. 2x + 1

and a = 1 to see that the area is inﬁnite. √

uh a

50. (a) 1 ≤

+∞

1dx = +∞ 2

+∞ x dx 1 dx ≤ = lim − 3 = 1/24 →+∞ x5 + 1 x4 3x 2 2 +∞ +∞ xex xex dx dx ≥ ≥ = +∞ 2x + 1 2x + 1 2x +1 1 1

+∞

(b)

x3 + 1 for x ≥ 2, x

∞

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Exercise Set 8.8

3/2

t3/2 dt = lim

lim

x→+∞

≥

t3 dt

x→+∞

52. (b) u =

√

x→+∞

+∞

x, 0

2 (2x)5/2 , 5 0 +∞ = +∞ so 1 + t3 dt = +∞; by L’Hˆ opital’s Rule =

= lim

x5/2

x→+∞

2 (2x)5/2 5

1 + t3 dt

lim

2 dt = t5/2 5

1 + (2x)3 2 = lim 3/2 x→+∞ (5/2)x

√ 1/x3 + 8 = 8 2/5 5/2

51.

us uf

√ +∞ +∞ cos x √ dx = 2 cos u du; cos u du diverges by Part (a). x 0 0

363

dx 1 1 x 53. Let x = r tan θ to get cos θ dθ = 2 sin θ + C = √ = 2 +C 2 r r (r2 + x2 )3/2 r r2 + x2 2πN Ir x 2πN I √ lim (+/ r2 + +2 − a/ r2 + a2 ) lim so u = = →+∞ r 2 r 2 + x2 a kr →+∞ k 2πN I = (1 − a/ r2 + a2 ). kr M to get 2RT

3/2 −2 M M 1 2RT 8RT 2 4 = =√ (a) v¯ = √ 2 2RT M πM π 2RT π

3/2 √ −5/2 M M 3 π 3RT 4 3RT 2 = (b) vrms = √ so vrms = 8 2RT M M π 2RT

54. Let a2 =

55. (a) Satellite’s weight = w(x) = k/x2 lb when x = distance from center of Earth; w(4000) = 6000 4000+ 10 so k = 9.6 × 10 and W = 9.6 × 1010 x−2 dx mi·lb. 4000

+∞

10 −2

9.6 × 10 x

(b)

dx = lim −9.6 × 10 /x 10

→+∞

4000

= 2.4 × 107 mi·lb

4000

1 −st 1 56. (a) L{1} = e dt = lim − e = →+∞ s s 0 0 +∞ +∞ 1 −(s−2)t 1 e e−st e2t dt = e−(s−2)t dt = lim − = (b) L{e2t } = →+∞ s−2 s−2 0 0 0 +∞ −st e 1 e−st sin t dt = lim 2 (c) L{sin t} = (−s sin t − cos t) = 2 →+∞ s + 1 s +1 0 0 +∞ e−st s e−st cos t dt = lim 2 (d) L{cos t} = (−s cos t + sin t) = 2 →+∞ s + 1 s +1 0 0

−st

uh a

+∞

57. (a) L{f (t)} =

→+∞

+∞

(b) L{f (t)} =

te−st dt = lim −(t/s + 1/s2 )e−st 2 −st

t e 0

+∞

= 0

1 s2 −st

dt = lim −(t /s + 2t/s + 2/s )e →+∞

1 e−st dt = lim − e−st →+∞ s

2 s3

−3s

= 3

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Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

58.

100

1000

10,000

0.8862269

Chapter 8

2 e−u du = π/a 0 0 +∞ +∞ √ √ 2 2 2 2 2 −x /2σ e dx = √ e−u du = 1 (b) x = 2σu, dx = 2σ du, √ π 2πσ 0 0

59. (a) u =

√

ax, du =

√

+∞

a dx, 2

−ax2

2 dx = √ a

√ 2 e−x dx ≈ 0.8862; π/2 ≈ 0.8862 0 3 +∞ +∞ −x2 −x2 −x2 e dx so E = e dx = e dx + (b)

+∞

xe−x dx =

dx < 3

1 −9 e < 7 × 10−5 2

1 dx ≈ 1.047; π/3 ≈ 1.047 6+1 x 0 4 +∞ +∞ 1 1 1 dx = dx + dx so (b) 6+1 6+1 6+1 x x x 4 0 0 +∞ +∞ 1 1 1 dx < dx = < 2 × 10−4 E= 6+1 6 5 x x 5(4) 4 4

61. (a)

+∞ −x2

60. (a)

us uf

364

+∞

→+∞

if p = 0, then 0

63. If p = 1, then

if p = 1, then 0

1 epx dx = lim epx →+∞ p

1 p (e − 1) = →+∞ p

= lim

+∞

−1/p, +∞,

p<0 . p>0

1 dx = lim+ ln x = +∞; x →0 1−p 1 dx x 1/(1 − p), p < 1 1−p = lim = lim [(1 − + )/(1 − p)] = . +∞, p>1 xp →0+ 1 − p →0+

(1)dx = lim x = +∞,

62. If p = 0, then

√ 64. u = 1 − x, u2 = 1 − x, 2u du = −dx; 1 0 √ 1 √ 2 2 − u du = 2 2 − u2 du = u 2 − u2 + 2 sin−1 (u/ 2) = 2 + π/2 −2

cos(u2 )du ≈ 1.809

65. 2

sin(1 − u2 )du = 2

66. −2 1

sin(1 − u2 )du ≈ 1.187 0

uh a

CHAPTER 8 SUPPLEMENTARY EXERCISES

1. (a) (c) (e) (g) (i)

integration by parts, u = x, dv = sin x dx reduction formula u-substitution: u = x3 + 1 integration by parts: dv = dx, u = tan−1 x u-substitution: u = 4 − x2

(b) (d) (f ) (h)

u-substitution: u = sin x u-substitution: u = tan x u-substitution: u = x + 1 trigonometric substitution: x = 2 sin θ

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Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Chapter 8 Supplementary Exercises

5. (a) #40 (d) #108

(b) #57 (e) #52

√ 0

x3 x2 + 1

x x2

dx, du = 2x dx, v =

1 2 2 dx = x x + 1 − 2 0

√

2 2 − (x2 + 1)3/2 3

√

x2 + 1;

x(x2 + 1)1/2 dx

1 = 0

√

6. (a) u = x2 , dv = √

sin θ tan θ

(b) x = 3 sin θ √ (e) x = 3 tan θ

us uf

1 3 1 9

2. (a) x = 3 tan θ (d) x = 3 sec θ

365

√ 2 √ 2 − [2 2 − 1] = (2 − 2)/3 3

(b) u2 = x2 + 1, x2 = u2 − 1, 2x dx = 2u du, x dx = u du; 1 1 √2 2 x3 x2 u −1 √ √ dx = x dx = u du 2 2 u x +1 x +1 0 1 0 √2 √2 √ 1 3 2 (u − 1)du = = (2 − 2)/3 u −u = 3 1 1

7. (a) u = 2x, 1 1 1 4 4 sin u du = − sin3 u cos u + sin 2x dx = 2 2 4 1 1 3 − sin u cos u + = − sin3 u cos u + 8 8 2

3 4

1 2

sin u du

1 3 3 = − sin3 u cos u − sin u cos u + u + C 16 16 8

3 3 1 sin 2x cos 2x + x + C = − sin3 2x cos 2x − 8 16 8

1 1 1 sin u − sin3 u + sin5 u + C 2 3 10

1 1 1 sin(x2 ) − sin3 (x2 ) + sin5 (x2 ) + C 2 3 10

(b) u = x2 , 1 1 (cos u)(1 − sin2 u)2 du cos5 u du = x cos5 (x2 )dx = 2 2 1 1 = cos u du − cos u sin2 u du + cos u sin4 u du 2 2

uh a

(b) With x = sin θ: 1 1 dx = − dθ = − 2 csc 2θ dθ x3 − x sin θ cos θ = − ln | csc 2θ − cot 2θ| + C = ln | cot θ| + C = ln

√

1 − x2 + C, 0 < |x| < 1. |x|

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Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Chapter 8

us uf

(c) By partial fractions: 1/2 1/2 1 1 1 + dx = − ln |x| + ln |x + 1| + ln |x − 1| + C − + x x+1 x−1 2 2 |x2 − 1| = ln + C; valid for all x except x = 0, ±1. |x|

366

√ 9. (a) With u = x: √ 1 1 du = 2 sin−1 (u/ 2) + C = 2 sin−1 ( x/2) + C; dx = 2 √ √ √ x 2−x 2 − u2 √ with u = 2 − x: √ √ √ 1 1 du = −2 sin−1 (u/ 2) + C = −2 sin−1 ( 2 − x/ 2) + C; dx = −2 √ √ √ x 2−x 2 − u2 completing the square: 1 dx = sin−1 (x − 1) + C. 1 − (x − 1)2

(b) In the three results in Part (a) the antiderivatives diﬀer by a constant, in particular √ √ 2 sin−1 ( x/2) = π − 2 sin−1 ( 2 − x/ 2) = π/2 + sin−1 (x − 1). 2

3−x 3−x C A B dx, 2 = + 2+ ; A = −4, B = 3, C = 4 3 + x2 (x + 1) x x x x x +1 1 2 3 A = −4 ln |x| − + 4 ln |x + 1| x 1 3 3 3 3 + 4 ln 3) − (−4 ln 1 − 3 + 4 ln 2) = − 8 ln 2 + 4 ln 3 = + 4 ln 2 2 2 4

= (−4 ln 2 −

10. A =

11. Solve y = 1/(1 + x2 ) for x to get

1−y and integrate with respect to x= y 1

1−y y to get A = dy (see ﬁgure) y 0

+∞

13. V = 2π

+∞

−x

12. A =

ln x − 1 ln x dx = lim − →+∞ x2 x

→+∞

but lim e− (+ + 1) = lim →+∞

+∞

15. u = cos θ, −

= 1/e e

(x + 1) = 2π lim 1 − e− (+ + 1) 0

→+∞

++1 1 = lim = 0 so V = 2π →+∞ e e

dx 1 1 π −1 tan tan−1 (+/a) = = 1, a = π/2 = lim (x/a) = lim →+∞ a →+∞ x2 + a2 a 2a 0

uh a

14.

→+∞

−x

dx = 2π lim −e

y = 1 / (1 + x2)

2 u1/2 du = − cos3/2 θ + C 3

16. Use Endpaper Formula (31) to get

tan7 θdθ =

1 1 1 tan6 θ − tan4 θ + tan2 θ + ln | cos θ| + C. 6 4 2

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Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Chapter 8 Supplementary Exercises

u2 du =

1 tan3 (x2 ) + C 6

π/2

π/2 π/2 3 1 cos3 θ sin θ + cos2 θ dθ 4 4 −π/2 −π/2 π/2 3 1 π/2 3 1 1 3π = √ + dθ = √ π = √ cos θ sin θ 2 −π/2 4 2 2 4 22 8 2 −π/2

1 cos4 θ dθ = √ 2 −π/2

1 √ 2

√ √ 18. x = (1/ 2) sin θ, dx = (1/ 2) cos θ dθ,

√ 1 1 du = √ tan−1 [(sin θ − 3)/ 3] + C +3 3

20.

cos θ dθ, let u = sin θ − 3, (sin θ − 3)2 + 3

x+3 dx, let u = x + 1, (x + 1)2 + 1 u+2 2 2 −1/2 √ +√ du = u(u + 1) du = u2 + 1 + 2 sinh−1 u + C u2 + 1 u2 + 1 = x2 + 2x + 2 + 2 sinh−1 (x + 1) + C

21.

√ √ 19. x = 3 tan θ, dx = 3 sec2 θ dθ, x 1 1 1 1 cos θ dθ = sin θ + C = √ +C dθ = 3 3 sec θ 3 3 3 + x2

us uf

1 17. u = tan(x ), 2 2

x2 + 2x + 2 + 2 ln( x2 + 2x + 2 + x + 1) + C.

1 dx. x3 − x2

22. Let x = tan θ to get

Alternate solution: let x + 1 = tan θ, (tan θ + 2) sec θ dθ = sec θ tan θ dθ + 2 sec θ dθ = sec θ + 2 ln | sec θ + tan θ| + C =

uh a

A C 1 B = + 2+ ; A = −1, B = −1, C = 1 so x2 (x − 1) x x x−1 1 1 1 1 dx + dx − dx = − ln |x| + + ln |x − 1| + C − x x2 x−1 x x − 1 1 + C = cot θ + ln tan θ − 1 + C = cot θ + ln |1 − cot θ| + C = + ln x x tan θ

23.

367

1 A B C 1 1 1 = + + ;A=− ,B= ,C= so (x − 1)(x + 2)(x − 3) x−1 x+2 x−3 6 15 10 1 1 1 1 1 1 − dx + dx + dx x−1 15 x+2 10 x−3 6 1 1 1 ln |x + 2| + ln |x − 3| + C = − ln |x − 1| + 6 15 10

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Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH x(x2

A Bx + C 1 = + 2 ; A = 1, B = C = −1 so + x + 1) x x +x+1

−x − 1 dx = − 2 x +x+1

x+1 dx = − (x + 1/2)2 + 3/4

u + 1/2 du, u2 + 3/4

u = x + 1/2

so 25. u = 2 0

2x + 1 1 dx 1 +C = ln |x| − ln(x2 + x + 1) − √ tan−1 √ 2 x(x2 + x + 1) 3 3 √

x − 4, x = u2 + 4, dx = 2u du, 2 2 4 2u2 −1 du = 2 du = 2u − 4 tan (u/2) = 4 − π 1− 2 u +4 u2 + 4 0 0

√ 1 1 = − ln(u2 + 3/4) − √ tan−1 (2u/ 3) + C1 2 3

24.

Chapter 8

us uf

368

2 du = u2 − 1

√ x 1 1 e +1−1 √ − du = ln |u − 1| − ln |u + 1| + C = ln +C u−1 u+1 ex + 1 + 1

2u du, +1 1 1 1 2u2 π −1 du = (2u − 2 tan u) = 2 − 1− 2 du = 2 + 1 u2 + 1 u 2 0 0 ex − 1, ex = u2 + 1, x = ln(u2 + 1), dx =

lim −

→+∞

1 2 2(x + 1)

1 bx lim tan−1 →+∞ ab a

= lim

→+∞

−

= lim

1 1 1 = + 2 2 2 2(+ + 1) 2(a + 1) 2(a + 1)

1 b+ π tan−1 = ab a 2ab

30.

√

2u du, −1

28. u =

29.

ex + 1, ex = u2 − 1, x = ln(u2 − 1), dx =

√

27. u =

√ 26. u = x, x = u2 , dx = 2u du, 3 3 3 9 u2 3 −1 u du = 2u − 6 tan 2 1 − =6− π du = 2 2+9 2+9 u u 3 2 0 0 0

1 4

31. Let u = x4 to get

→+∞

√

1 1 1 du = sin−1 u + C = sin−1 (x4 ) + C. 4 4 1 − u2

(cos32 x sin30 x − cos30 x sin32 x)dx =

uh a

32.

cos30 x sin30 x(cos2 x − sin2 x)dx

1 = 30 2

sin30 2x cos 2x dx =

sin31 2x +C 31(231 )

√ √ √ 2 1 2 33. x − x − 4dx = √ ( x + 2 − x − 2)dx = [(x + 2)3/2 − (x − 2)3/2 ] + C 3 2

34.

1 1 dx = − x10 (1 + x−9 ) 9

1 1 1 du = − ln |u| + C = − ln |1 + x−9 | + C u 9 9

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Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Chapter 8 Supplementary Exercises

−1

+∞

−t

e dt = lim −e

36. (a) Γ(1) =

→+∞

+∞

(b) Γ(x + 1) =

2 3 x−2 3 + − − x + 4 x − 5 x2 + 1 (x2 + 1)2 1 3 x − ln(x2 + 1) − 2 2

x −1 + tan x + C x2 + 1

= lim (−e− + 1) = 1 →+∞

tx e−t dt; let u = tx , dv = e−t dt to get

0 x −t

+∞

Γ(x + 1) = −t e

+∞

x−1 −t

x −t

+∞

e dt = −t e

us uf

(b) −

B Cx + D Ex + F A + + 2 + 2 x+4 x−5 x +1 (x + 1)2

35. (a) (x + 4)(x − 5)(x2 + 1)2 ;

+ xΓ(x) 0

tx = 0 (by multiple applications of L’Hˆ opital’s rule) t→+∞ et t→+∞ so Γ(x + 1) = xΓ(x)

lim tx e−t = lim

It appears that Γ(n) = (n − 1)! if n is a positive integer. +∞ +∞ √ √ √ 2 1 −1/2 −t (d) Γ = t e dt = 2 e−u du (with u = t) = 2( π/2) = π 2 0 0 1√ 3 3 3√ 1 1 3 5 = = Γ = = Γ (e) Γ π, Γ π 2 2 2 2 2 4 2 2

37. (a) t = − ln x, x = e−t , dx = −e−t dt, 1 0 (ln x)n dx = − (−t)n e−t dt = (−1)n 0

+∞ n

1/n

+∞

tn e−t dt = (−1)n Γ(n + 1)

1/n−1

(b) t = x , x = t , dx = (1/n)t dt, +∞ +∞ n e−x dx = (1/n) t1/n−1 e−t dt = (1/n)Γ(1/n) = Γ(1/n + 1) cos θ − cos θ0 = 2 sin2 (θ0 /2) − sin2 (θ/2) = 2(k 2 − k 2 sin2 φ) = 2k 2 cos2 φ √ 1 1 = 2 k cos φ; k sin φ = sin(θ/2) so k cos φ dφ = cos(θ/2) dθ = 1 − sin2 (θ/2) dθ 2 2 1 2k cos φ = 1 − k 2 sin2 φ dθ, thus dθ = dφ and hence 2 1 − k 2 sin2 φ 8L π/2 1 L π/2 1 2k cos φ √ · dφ = 4 dφ T = 2 g 0 g 0 2k cos φ 1 − k 2 sin φ 1 − k 2 sin2 φ

uh a

38. (a)

369

(b) If L = 1.5 ft and θ0 = (π/180)(20) = π/9, then √ π/2 3 dφ ≈ 1.37 s. T = 2 0 2 1 − sin (π/18) sin2 φ

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370

Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Chapter 8

CHAPTER 8 HORIZON MODULE

us uf

1. The depth of the cut equals the terrain elevation minus the track elevation. From Figure 2, the 1 cross sectional area of a cut of depth D meters is 10D + 2 · D2 = D2 + 10D square meters. 2 Distance from Terrain elevation Track elevation Depth of cut Cross-sectional area town A (m) (m) (m) (m) f (x) of cut (m2 ) 100 101 102 103 104 105 106 107 108 109 110

0 56 96 119 0 11 336 375 416 551 600

0 4 6 7 0 1 14 15 16 19 20

100 105 108 110 104 106 120 122 124 128 130

0 2000 4000 6000 8000 10,000 12,000 14,000 16,000 18,000 20,000

2000

The total volume of dirt to be excavated, in cubic meters, is

f (x) dx.

By Simpson’s Rule, this is approximately

20,000 − 0 [0 + 4 · 56 + 2 · 96 + 4 · 119 + 2 · 0 + 4 · 11 + 2 · 336 + 4 · 375 + 2 · 416 + 4 · 551 + 600] 3 · 10 = 4,496,000 m3 .

Distance from Terrain elevation Track elevation Depth of cut Cross-sectional area town A (m) (m) f (x) of cut (m2 ) (m) (m) 130 135 139 142 145 147 148 146 143 139 133

110 109.8 109.6 109.4 109.2 109 108.8 108.6 108.4 108.2 108

20,000 20,100 20,200 20,300 20,400 20,500 20,600 20,700 20,800 20,900 21,000

2. (a)

4 · 4,496,000 = 17,984,000 dollars.

Excavation costs $4 per m3 , so the total cost of the railroad from kA to M is about

20 25.2 29.4 32.6 35.8 38 39.2 37.4 34.6 30.8 25

300 887.04 1158.36 1388.76 1639.64 1824 1928.64 1772.76 1543.16 1256.64 875

21,000

uh a

The total volume of dirt to be excavated, in cubic meters, is

f (x) dx. 20,000

By Simpson’s Rule this is approximately

21,000 − 20,000 [600 + 4 · 887.04 + 2 · 1158.36 + . . . + 4 · 1256.64 + 875] = 1,417,713.33 m3 . 3 · 10

The total cost of a trench from M to N is about 4 · 1,417,713.33 ≈ 5,670,853 dollars.

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Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Chapter 8 Horizon Module

Distance from Terrain elevation Track elevation Depth of cut Cross-sectional area town A (m) (m) (m) (m) f (x) of cut (m2 ) 108 106 104 102 100 98 96 94 92 90 88

875 336 24 96 96 0 56 144 56 11 0

25 14 2 6 6 0 4 8 4 1 0

133 120 106 108 106 98 100 102 96 91 88

us uf

21,000 22,000 23,000 24,000 25,000 26,000 27,000 28,000 29,000 30,000 31,000

(b)

371

31,000

f (x) dx. By Simpson’s

The total volume of dirt to be excavated, in cubic meters, is

21,000

Rule this is approximately

31,000 − 21,000 [875 + 4 · 336 + 2 · 24 + . . . + 4 · 11 + 0] = 1,229,000 m3 . 3 · 10

The total cost of the railroad from N to B is about 4 · 1,229,000 ≈ 4,916,000 dollars.

3. The total cost if trenches are used everywhere is about 17,984,000 + 5,670,853 + 4,916,000 = 28,570,853 dollars.

1 2 π5 ≈ 119.27 m2 . The length of the 2 tunnel is 1000 m, so the volume of dirt to be removed is about 1000AT ≈ 1,119,269.91 m3 , and the drilling and dirt-piling costs are 8 · 1000AT ≈ 954,159 dollars. (b) To extend the tunnel from a length of x meters to a length of x + dx meters, we must move a volume of AT dx cubic meters of dirt a distance of about x meters. So the cost of this extension is about 0.06 × AT dx dollars. The cost of moving all of the dirt in the tunnel is therefore 1000 1000 x2 0.06 × AT dx = 0.06AT = 30,000AT ≈ 3,578,097 dollars. 2 0 0

4. (a) The cross-sectional area of a tunnel is AT = 80 +

uh a

5. The total cost of the railroad, using a tunnel, is 17,894,000 + 4,532,257 + 4,916,000 + 27,432,257 dollars, which is smaller than the cost found in Exercise 3. It will be cheaper to build the railroad if a tunnel is used.

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Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH CHAPTER 9

us uf

Mathematical Modeling with Differential Equations EXERCISE SET 9.1 3

1. y = 2x2 ex

= x2 y and y(0) = 2 by inspection.

2. y = x3 − 2 sin x, y(0) = 3 by inspection.

dy dy = c; (1 + x) = (1 + x)c = y dx dx (b) second order; y = c1 cos t − c2 sin t, y + y = −c1 sin t − c2 cos t + (c1 sin t + c2 cos t) = 0

3. (a) ﬁrst order;

c dy + y = 2 − e−x/2 + 1 + ce−x/2 + x − 3 = x − 1 dx 2 t (b) second order; y = c1 e − c2 e−t , y − y = c1 et + c2 e−t − c1 et + c2 e−t = 0

6. 2x + y 2 + 2xy

dy = 0, by inspection. dx

dy y2 dy 1 dy dy + y, (1 − xy) = y 2 , = =x 1 − xy dx dx dx y dx

4. (a) ﬁrst order; 2

d 3x ye = 0, ye3x = C, y = Ce−3x dx dy separation of variables: = −3dx, ln |y| = −3x + C1 , y = ±e−3x eC1 = Ce−3x y including C = 0 by inspection

= e3x ,

7. (a) IF: µ = e3

d [ye−2t ] = 0, ye−2t = C, y = Ce2t dt dy = 2dt, ln |y| = 2t + C1 , y = ±eC1 e2t = Ce2t separation of variables: y including C = 0 by inspection dt

= e−2t ,

(b) IF: µ = e−2

8. (a) IF: µ = e−4

x dx

= e−2x ,

2 d −2x2 ye = 0, y = Ce2x dx

2 2 dy = 4x dx, ln |y| = 2x2 + C1 , y = ±eC1 e2x = Ce2x y including C = 0 by inspection d t ye = 0, y = Ce−t (b) IF: µ = e dt = et , dt dy = −dt, ln |y| = −t + C1 , y = ±eC1 e−t = Ce−t separation of variables: y including C = 0 by inspection

9. µ = e 3dx = e3x , e3x y = ex dx = ex + C, y = e−2x + Ce−3x

uh a

separation of variables:

10. µ = e2

x dx

= ex ,

2 2 2 d x2 1 2 1 ye = xex , yex = ex + C, y = + Ce−x dx 2 2

372

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Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Exercise Set 9.1

13.

14.

=e ,e y=

ex cos(ex )dx = sin(ex ) + C, y = eâ&#x2C6;&#x2019;x sin(ex ) + Ceâ&#x2C6;&#x2019;x

dy 1 + 2y = , Âľ = e 2dx = e2x , e2x y = dx 2

1 2x 1 1 e dx = e2x + C, y = + Ceâ&#x2C6;&#x2019;2x 2 4 4

2 2 1 x dy + 2 y = 0, Âľ = e (x/(x +1))dx = e 2 ln(x +1) = x2 + 1, dx x + 1 d 2 C y x + 1 = 0, y x2 + 1 = C, y = â&#x2C6;&#x161; dx x2 + 1 1 dy +y = , Âľ = e dx = ex , ex y = x dx 1+e

12.

11. Âľ = e

us uf

ex dx = ln(1 + ex ) + C, y = eâ&#x2C6;&#x2019;x ln(1 + ex ) + Ceâ&#x2C6;&#x2019;x 1 + ex

1 y 1

dy = dx, ln |y| = ln |x| + C1 , ln = C1 , = ÂąeC1 = C, y = Cx y x x x including C = 0 by inspection

16.

dy 1 = x2 dx, tanâ&#x2C6;&#x2019;1 y = x3 + C, y = tan 2 1+y 3

17.

â&#x2C6;&#x161; â&#x2C6;&#x161; x dy 2 2 dx, ln |1 + y| = â&#x2C6;&#x2019; 1 + x2 + C1 , 1 + y = Âąeâ&#x2C6;&#x2019; 1+x eC1 = Ceâ&#x2C6;&#x2019; 1+x , = â&#x2C6;&#x2019;â&#x2C6;&#x161; 2 1+y 1+x

sin x dx = sec x tan x dx, ey = sec x + C, y = ln(sec x + C) cos2 x

dy x2 â&#x2C6;&#x2019;1 = (1 + x) dx, tan y = x + + C, y = tan(x + x2 /2 + C) 1 + y2 2

y â&#x2C6;&#x2019; 1

1 dy dx 1

+ dy = csc x dx, ln = , â&#x2C6;&#x2019;

y = ln | csc x â&#x2C6;&#x2019; cot x| + C1 , y2 â&#x2C6;&#x2019; y sin x y yâ&#x2C6;&#x2019;1 1 yâ&#x2C6;&#x2019;1 = ÂąeC1 (csc x â&#x2C6;&#x2019; cot x) = C(csc x â&#x2C6;&#x2019; cot x), y = , C = 0; y 1 â&#x2C6;&#x2019; C(csc x â&#x2C6;&#x2019; cot x) by inspection, y = 0 is also a solution, as is y = 1.

uh a

23.

dy = ex dx, ln |y| + y 2 /2 = ex + C; by inspection, y = 0 is also a solution

2 2 dy = â&#x2C6;&#x2019;x dx, ln |y| = â&#x2C6;&#x2019;x2 /2 + C1 , y = ÂąeC1 eâ&#x2C6;&#x2019;x /2 = Ceâ&#x2C6;&#x2019;x /2 , including C = 0 by inspection y

21. ey dy =

22.

20.

1 +y y

1 x3 dx y 2 4 2 4 = ln(1 + x , ) + C , 2y = ln(1 + x ) + C, y = Âą [ln(1 + x4 ) + C]/2 1 1 + x4 2 4

18. y dy =

19.

â&#x2C6;&#x2019; 1, C = 0

1 3 x +C 3

â&#x2C6;&#x161; 1+x2

15.

y = Ceâ&#x2C6;&#x2019;

24.

373

1 3 cos y dy = dx, dy = 3 cos x dx, ln | sin y| = 3 sin x + C1 , tan y sec x sin y

sin y = Âąe3 sin x+C1 = ÂąeC1 e3 sin x = Ce3 sin x , C = 0, y = sinâ&#x2C6;&#x2019;1 Ce3 sin x , as is y = 0 by inspection

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Chapter 9

1 1 d dy + y = 1, Âµ = e (1/x)dx = eln x = x, [xy] = x, xy = x2 + C, y = x/2 + C/x dx x dx 2

1 3 + C, C = , y = x/2 + 3/(2x) 2 2

(a) 2 = y(1) =

25.

Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

us uf

374

(b) 2 = y(â&#x2C6;&#x2019;1) = â&#x2C6;&#x2019;1/2 â&#x2C6;&#x2019; C, C = â&#x2C6;&#x2019;5/2, y = x/2 â&#x2C6;&#x2019; 5/(2x) 2 2 dy x2 = x dx, ln |y| = + C1 , y = Â±eC1 ex /2 = Cex /2 y 2

1 2 1 = y(0) = C, so y = ex /2 2 2

27. Âµ = e

x dx

â&#x2C6;&#x2019;x2 /2

=e 2

y = â&#x2C6;&#x2019;1 + Cex

28. Âµ = e

xeâ&#x2C6;&#x2019;x

dx = â&#x2C6;&#x2019;eâ&#x2C6;&#x2019;x 2

=e,ey=

29. (y + cos y) dy = 4x2 dx, C=

, 3 = â&#x2C6;&#x2019;1 + C, C = 4, y = â&#x2C6;&#x2019;1 + 4ex

â&#x2C6;&#x2019;x2 /2

+ C,

â&#x2C6;&#x2019;

(b)

2et dt = 2et + C, y = 2 + Ceâ&#x2C6;&#x2019;t , 1 = 2 + C, C = â&#x2C6;&#x2019;1, y = 2 â&#x2C6;&#x2019; eâ&#x2C6;&#x2019;t y2 Ï&#x20AC;2 Ï&#x20AC;2 4 4 4 + sin y = x3 + C, + sin Ï&#x20AC; = (1)3 + C, = + C, 2 3 2 3 2 3

4 Ï&#x20AC;2 â&#x2C6;&#x2019; , 3y 2 + 6 sin y = 8x3 + 3Ï&#x20AC; 2 â&#x2C6;&#x2019; 8 3 2

dy 1 = (x + 2)ey , eâ&#x2C6;&#x2019;y dy = (x + 2)dx, â&#x2C6;&#x2019;eâ&#x2C6;&#x2019;y = x2 + 2x + C, â&#x2C6;&#x2019;1 = C, 2 dx 1 1 1 â&#x2C6;&#x2019;eâ&#x2C6;&#x2019;y = x2 + 2x â&#x2C6;&#x2019; 1, eâ&#x2C6;&#x2019;y = â&#x2C6;&#x2019; x2 â&#x2C6;&#x2019; 2x + 1, y = â&#x2C6;&#x2019; ln 1 â&#x2C6;&#x2019; 2x â&#x2C6;&#x2019; x2 2 2 2

30.

(a) 1 = y(0) = C so C = 1, y = ex

26.

31. 2(y â&#x2C6;&#x2019; 1) dy = (2t + 1) dt, y 2 â&#x2C6;&#x2019; 2y = t2 + t + C, 1 + 2 = C, C = 3, y 2 â&#x2C6;&#x2019; 2y = t2 + t + 3 sinh x y = cosh x, Âµ = e (sinh x/ cosh x)dx = eln cosh x = cosh x, cosh x

1 1 1 2 (cosh x)y = cosh x dx = (cosh 2x + 1)dx = sinh 2x + x + C = 2 4 2 1 1 1 1 1 y = sinh x + x sech x + C sech x, = C, y = sinh x + x sech x + 2 4 2 2 2

32. y +

dy dx 1 = , ln |y| = ln |x| + C1 , y 2x 2

33. (a)

x = â&#x20AC;&#x201C;0.5y2 2

y 2

x = â&#x20AC;&#x201C;1.5y

x = y2

uh a

|y| = C|x|1/2 , y 2 = Cx;

by inspection y = 0 is also a solution.

1 1 sinh x cosh x + x + C, 2 2 1 sech x 4

x = â&#x20AC;&#x201C; 3y2

x = 2y2

y=0 -2

x 2

(b) 1 = C(2)2 , C = 1/4, y 2 = x/4 -2

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x = 2.5y2

Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Exercise Set 9.1

34. (a) y dy = â&#x2C6;&#x2019;x dx,

y = â&#x2C6;&#x161;9 â&#x20AC;&#x201C; x 2 y = â&#x2C6;&#x161;2.25 â&#x20AC;&#x201C; x 2

25 â&#x2C6;&#x2019; x2

â&#x2C6;&#x161;

y 3

y = â&#x2C6;&#x161;0.25 â&#x20AC;&#x201C; x 2

us uf

(b) y =

y2 x2 = â&#x2C6;&#x2019; + C1 , y = Âą C 2 â&#x2C6;&#x2019; x2 2 2

375

-3

y = â&#x20AC;&#x201C; â&#x2C6;&#x161;1 â&#x20AC;&#x201C; x 2

y = â&#x20AC;&#x201C; â&#x2C6;&#x161;4 â&#x20AC;&#x201C; x 2

-3

36. y + 2y = 3et , Âľ = e2

-2

37. (1 â&#x2C6;&#x2019; y 2 ) dy = x2 dx,

38.

x3 y3 = + C1 , x3 + y 3 â&#x2C6;&#x2019; 3y = C 3 3

x 2

1 +y y

yey

dy = dx, ln |y| +

y2 = x + C1 , 2

= ÂąeC1 ex = Cex including C = 0 y 3 x -5

5 -3

-2

-100

-1

C = â&#x20AC;&#x201C;1 C = â&#x20AC;&#x201C;2

2 C = â&#x20AC;&#x201C;1 C = â&#x20AC;&#x201C;2

yâ&#x2C6;&#x2019;

= e2t ,

100 C=2 C=1 C=0 C=2 C=1

-2

d 2t ye = 3e3t , ye2t = e3t + C, dt y = et + Ceâ&#x2C6;&#x2019;2t

1.5

C=0

x dx dy =â&#x2C6;&#x2019; 2 , y x +4 1 ln |y| = â&#x2C6;&#x2019; ln(x2 + 4) + C1 , 2 C y=â&#x2C6;&#x161; x2 + 4

35.

y = â&#x20AC;&#x201C; â&#x2C6;&#x161;6.25 â&#x20AC;&#x201C; x 2

uh a

1 1 and thus never through (0, 0). 39. Of the solutions y = 2 , all pass through the point 0, â&#x2C6;&#x2019; 2x â&#x2C6;&#x2019; C C A solution of the initial value problem with y(0) = 0 is (by inspection) y = 0. The methods of Example 4 fail because the integrals there become divergent when the point x = 0 is included in the integral.

1 1 40. If y0 = 0 then, proceeding as before, we get C = 2x2 â&#x2C6;&#x2019; , C = 2x20 â&#x2C6;&#x2019; , and y y0 1 y= , which is deďŹ ned for all x provided 2x2 is never equal to 2x20 â&#x2C6;&#x2019; 1/y0 ; this 2x2 â&#x2C6;&#x2019; 2x20 + 1/y0 last condition will be satisďŹ ed if and only if 2x20 â&#x2C6;&#x2019; 1/y0 < 0, or 0 < 2x20 y0 < 1. If y0 = 0 then y = 0 is, by inspection, also a solution for all real x.

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376

Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Chapter 9

dy x2 = xey , e−y dy = x dx, −e−y = + C, x = 2 when y = 0 so −1 = 2 + C, C = −3, x2 + 2e−y = 6 dx 2

42.

3x2 dy = , 2y dy = 3x2 dx, y 2 = x3 + C, 1 = 1 + C, C = 0, dx 2y

us uf

41.

0 0

dy = rate in − rate out, where y is the amount of salt at time t, dt y dy 1 dy 1 (2) = 8 − y, so + y = 8 and y(0) = 25. = (4)(2) − dt 50 25 dt 25

µ = e (1/25)dt = et/25 , et/25 y = 8et/25 dt = 200et/25 + C,

y = 200 + Ce−t/25 , 25 = 200 + C, C = −175, (a) y = 200 − 175e−t/25 oz

(b) when t = 25, y = 200 − 175e−1 ≈ 136 oz

y 1 dy 1 dy = (5)(10) − (10) = 50 − y, so + y = 50 and y(0) = 0. dt 200 20 dt 20

1 µ = e 20 dt = et/20 , et/20 y = 50et/20 dt = 1000et/20 + C,

44.

1.6

43.

y 2 = x3 , y = x3/2 passes through (1, 1).

y = 1000 − 1000e−t/20 lb

(b)

when t = 30, y = 1000 − 1000e−1.5 ≈ 777 lb

(a)

y = 1000 + Ce−t/20 , 0 = 1000 + C, C = −1000;

45. The volume V of the (polluted) water is V (t) = 500 + (20 − 10)t = 500 + 10t; if y(t) is the number of pounds of particulate matter in the water, dt dy y 1 dy 1 then y(0) = 50, and = 0 − 10 = − y, + y = 0; µ = e 50+t = 50 + t; dt V 50 + t dt 50 + t d [(50 + t)y] = 0, (50 + t)y = C, 2500 = 50y(0) = C, y(t) = 2500/(50 + t). dt The tank reaches the point of overﬂowing when V = 500 + 10t = 1000, t = 50 min, so y = 2500/(50 + 50) = 25 lb.

46. The volume of the lake (in gallons) is V = 264πr2 h = 264π(15)2 3 = 178,200π gals. Let y(t) denote y dy y = 0 − 103 the number of pounds of mercury salts at time t, then = − lb/h and V 178.2π dt dy dt t y0 = 10−5 V = 1.782π lb; =− , ln y = − + C1 , y = Ce−t/(178.2π) , and y 178.2π 178.2π

uh a

C = y(0) = y0 10−5 V = 1.782π, y = 1.782πe−t/(178.2π) lb of mercury salts.

1 2 3 4 5 6 7 8 9 10 11 12 t y(t) 5.588 5.578 5.568 5.558 5.549 5.539 5.529 5.519 5.509 5.499 5.489 5.480

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Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Exercise Set 9.1

377

c d ct/m gm ct/m dv + v = −g, µ = e(c/m) dt = ect/m , ve e = −gect/m , vect/m = − + C, dt m dt c gm gm gm gm −ct/m gm +Ce−ct/m , but v0 = v(0) = − +C, C = v0 + ,v = − + v0 + e v=− c c c c c mg with vτ and −ct/m with −gt/vτ in (23). (b) Replace c vτ (c) From Part (b), s(t) = C − vτ t − (v0 + vτ ) e−gt/vτ ; g vτ vτ vτ s0 = s(0) = C −(v0 +vτ ) , C = s0 +(v0 +vτ ) , s(t) = s0 −vτ t+ (v0 +vτ ) 1 − e−gt/vτ g g g

us uf

47. (a)

48. Given m = 240, g = 32, vτ = mg/c: with a closed parachute vτ = 120 so c = 64, and with an open parachute vτ = 24, c = 320.

(a) Let t denote time elapsed in seconds after the moment of the drop. From Exercise 47(b), while the parachute is closed v(t) = e−gt/vτ (v0 + vτ ) − vτ = e−32t/120 (0 + 120) − 120 = 120 e−4t/15 − 1 and thus −20/3 − 1 ≈ −119.85, so the parachutist is falling at a speed of 119.85 ft/s v(25) = 120 e 120 120 1 − e−4t/15 , when the parachute opens. From Exercise 47(c), s(t) = s0 − 120t + 32 s(25) = 10000 − 120 · 25 + 450 1 − e−20/3 ≈ 7449.43 ft.

R d V (t) Rt/L dI V (t) + I= , , µ = e(R/L) dt = eRt/L , (eRt/L I) = e dt L L dt L

t 1 t 1 IeRt/L = I(0) + V (u)eRu/L du, I(t) = I(0)e−Rt/L + e−Rt/L V (u)eRu/L du. L 0 L 0 t

t 6 6 1 1 − e−5t/2 A. 12e5u/2 du = e−5t/2 e5u/2 = (a) I(t) = e−5t/2 5 5 4 0 0

lim I(t) =

t→+∞

6 A 5

(b)

49.

(b) If t denotes time elapsed after the parachute opens, then, by Exercise 47(c), 24 (−119.85 + 24) 1 − e−32t/24 = 0, with the solution (Newton’s s(t) = 7449.43 − 24t + 32 Method) t = 307.4 s, so the sky diver is in the air for about 25 + 307.4 = 332.4 s.

50. From Exercise 49 and Endpaper Table #42, t

t e2u 1 (2 sin u − cos u) 3e2u sin u du = 15e−2t + e−2t I(t) = 15e−2t + e−2t 3 5 0 0

1 1 = 15e−2t + (2 sin t − cos t) + e−2t . 5 5 dv ck = − g, v = −c ln(m0 − kt) − gt + C; v = 0 when t = 0 so 0 = −c ln m0 + C, dt m0 − kt m0 − gt. C = c ln m0 , v = c ln m0 − c ln(m0 − kt) − gt = c ln m0 − kt

uh a

51. (a)

(b) m0 − kt = 0.2m0 when t = 100 so m0 − 9.8(100) = 2500 ln 5 − 980 ≈ 3044 m/s. v = 2500 ln 0.2m0

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Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Chapter 9

dv dx dv dv dv dv = = v so m = mv . dt dx dt dx dt dx m mv dv = −dx, ln(kv 2 + mg) = −x + C; v = v0 when x = 0 so (b) kv 2 + mg 2k

us uf

52. (a) By the chain rule,

m m m m kv02 + mg ln(kv02 + mg), ln(kv 2 + mg) = −x + ln(kv02 + mg), x = ln . 2k 2k 2k kv 2 + mg 2k

3.56 × 10−3 m kv02 + mg (7.3 × 10−6 )(988)2 + (3.56 × 10−3 )(9.8) = ln ln ≈ 1298 m mg 2(7.3 × 10−6 ) (3.56 × 10−3 )(9.8) 2k

xmax =

√ √ π dh = −0.025 h, √ dh = −0.025dt, 2π h = −0.025t + C; h = 4 when 53. (a) A(h) = π(1)2 = π, π dt h √ √ 0.025 t = 0, so 4π = C, 2π h = −0.025t + 4π, h = 2 − t, h ≈ (2 − 0.003979 t)2 . 2π (b) h = 0 when t ≈ 2/0.003979 ≈ 502.6 s ≈ 8.4 min.

54. (a) A(h) = 6 2 4 − (h − 2)2 = 12 4h − h2 ,

√ √ dh = −0.025 h, 12 4 − h dh = −0.025dt, 12 4h − h2 dt 3/2 −8(4 − h) = −0.025t + C; h = 4 when t = 0 so C = 0,

2√4 − (h − 2)2

h−2

(4 − h)3/2 = (0.025/8)t, 4 − h = (0.025/8)2/3 t2/3 ,

1 1 1 dv = −0.04v 2 , 2 dv = −0.04dt, − = −0.04t + C; v = 50 when t = 0 so − = C, dt v v 50 1 1 50 dx dx 50 − = −0.04t − , v = cm/s. But v = so = , x = 25 ln(2t + 1) + C1 ; v 50 2t + 1 dt dt 2t + 1

55.

8 (4 − 0)3/2 = 2560 s ≈ 42.7 min 0.025

(b) h = 0 when t =

h ≈ 4 − 0.021375t2/3 ft

x = 0 when t = 0 so C1 = 0, x = 25 ln(2t + 1) cm. √ √ dv 1 = −0.02 v, √ dv = −0.02dt, 2 v = −0.02t + C; v = 9 when t = 0 so 6 = C, dt v

56.

√ dx dx so = (3 − 0.01t)2 , 2 v = −0.02t + 6, v = (3 − 0.01t)2 cm/s. But v = dt dt 100 100 x=− (3 − 0.01t)3 + C1 ; x = 0 when t = 0 so C1 = 900, x = 900 − (3 − 0.01t)3 cm. 3 3

uh a

57. Diﬀerentiate to get

dP 1 [H(x) + C] where µ = eP (x) , = p(x), µ dx constant. Then dy 1 µ + p(x)y = H (x) − 2 [H(x) + C] + p(x)y = q − dx µ µ

58. (a) Let y =

2 dy = − sin x + e−x , y(0) = 1. dx

d H(x) = µq, and C is an arbitrary dx p [H(x) + C] + p(x)y = q µ

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Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Exercise Set 9.2

379

1 [H(x)+C] is a solution µ of the initial value problem with y(x0 ) = y0 . This shows that the initial value problem has a solution.

us uf

(b) Given the initial value problem, let C = µ(x0 )y0 −H(x0 ). Then y =

To show uniqueness, suppose u(x) also satisﬁes (5) together with u(x0 ) = y0 . Following the 1 arguments in the text we arrive at u(x) = [H(x) + C] for some constant C. The initial µ condition requires C = µ(x0 )y0 − H(x0 ), and thus u(x) is identical with y(x). dH dH dy dG = G (x). But = h(y) and = g(x), hence dy dx dy dx

59. Suppose that H(y) = G(x) + C. Then y(x) is a solution of (10).

60. (a) y = x and y = −x are both solutions of the given initial value problem.

(b) y dy = − x dx, y 2 = −x2 + C; but y(0) = 0, so C = 0. Thus y 2 = −x2 , which is

impossible.

61. Suppose I1 ⊂ I is an interval with I1 = I,and suppose Y (x) is deﬁned on I1 and is a solution of (5) there. Let x0 be a point of I1 . Solve the initial value problem on I with initial value y(x0 ) = Y (x0 ). Then y(x) is an extension of Y (x) to the interval I, and by Exercise 58(b) applied to the interval I1 , it follows that y(x) = Y (x) for x in I1 .

y(0) = 2

y(0) = 1

x 1

dy + y = 1, µ = e dx

= ex ,

x 2

-1

y(0) = –1

d [yex ] = ex , yex = ex + C, y = 1 + Ce−x dx

EXERCISE SET 9.2

(a) −1 = 1 + C, C = −2, y = 1 − 2e−x

(b) 1 = 1 + C, C = 0, y = 1

uh a

y(–1) = 0

-2

y(1) = 1

x 2

y(0) = –1

-10

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Chapter 9

ye−2x =

µ = e−2

d −2x ye = −xe−2x , dx

= e−2x ,

1 (2x + 1)e−2x + C, 4

1 (2x + 1) + Ce2x 4

(a) 1 = 3/4 + Ce2 , C = 1/(4e2 ), y = (b) −1 = 1/4 + C, C = −5/4, y =

dy − 2y = −x, dx

us uf

Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

1 1 (2x + 1) + e2x−2 4 4

5 1 (2x + 1) − e2x 4 4

1 1 (2x + 1) + e2x+2 4 4

lim y = 1

lim y =

x→+∞

+∞ −∞,

if y0 < 1/4

if y0 ≥ 1/4

380

9. (a) IV, since the slope is positive for x > 0 and negative for x < 0.

(b) VI, since the slope is positive for y > 0 and negative for y < 0. (c) V, since the slope is always positive.

(d) II, since the slope changes sign when crossing the lines y = ±1.

(e) I, since the slope can be positive or negative in each quadrant but is not periodic. III, since the slope is periodic in both x and y.

(f )

11. (a) y0 = 1, yn+1 = yn + (xn + yn )(0.2) = (xn + 6yn )/5

(c)

0 0 1

xn y(xn) abs. error perc. error

1 0.2 1.20

0 1 0 0

0.4 1.48

0.6 1.86

0.8 2.35

1.0 2.98

0.2 1.24 0.04

0.4 1.58 0.10

0.6 2.04 0.19

0.8 2.65 0.30

1.0 3.44 0.46

d −x ye = xe−x , dx ye−x = −(x + 1)e−x + C, 1 = −1 + C, C = 2, y = −(x + 1) + 2ex

(b) y − y = x, µ = e−x ,

n xn yn

0.4

0.6

x 0.8

0.2

uh a

12. h = 0.1, yn+1 = (xn + 11yn )/10

n xn yn

0 0 1.00

1 0.1 1.10

0.2 1.22

0.3 1.36

0.4 1.53

0.5 1.72

0.6 1.94

0.7 2.20

0.8 2.49

9 0.9

10 1.0

2.82

3.19

In Exercise 11, y(1) ≈ 2.98; in Exercise 12, y(1) ≈ 3.19; the true solution is y(1) ≈ 3.44; so the absolute errors are approximately 0.46 and 0.25 respectively.

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Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Exercise Set 9.2

0 0 1

1 0.5 1.50

2 1

yn /2 3

1.5 2.84

2.11

y 9

4 2

5 2.5 4.64

3.68

6 3 5.72

7 3.5 6.91

8 4

n xn yn

√

8.23

us uf

13. y0 = 1, yn+1 = yn +

381

14. y0 = 1, yn+1 = yn + (xn − yn2 )/4 n xn yn

0 0 1

1 0.25 0.75

0.50 0.67

0.75 0.68

1.00 0.75

1.25 0.86

1.50 0.99

1.75 1.12

2.00 1.24

2 1.5

1 0.5

0 0 1

1 0.5 1.42

2 1 1.92

3 1.5 2.39

1.5

4 2 2.73

t 3

16. y0 = 0, yn+1 = yn + e−yn /10 0 0 0

1 0.1 0.10

0.2 0.19

0.3 0.27

y 1

0.4 0.35

0.5 0.42

0.6 0.49

0.7 0.55

0.8 0.60

0.9 0.66

1.0 0.71

n tn yn

1 sin yn 2

x 1

15. y0 = 1, yn+1 = yn +

0.5

17. h = 1/5, y0 = 1, yn+1 = yn + 0 0 1.00

1 0.2 1.06

uh a

n tn yn

2 0.4 0.90

18. (a) By inspection,

t 1

1 cos(2πn/5) 5

0.6 0.74

0.8 0.80

1.0 1.00

2 dy = e−x and y(0) = 0. dx 2

(b) yn+1 = yn + e−xn /20 = yn + e−(n/20) /20 and y20 = 0.7625. From a CAS, y(1) = 0.7468.

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382

Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Chapter 9

19. (b) y dy = â&#x2C6;&#x2019;x dx, y 2 /2 = â&#x2C6;&#x2019;x2 /2 + C1 , x2 + y 2 = C; if y(0) = 1 then C = 1 so y(1/2) =

us uf

dy 1 â&#x2C6;&#x161; â&#x2C6;&#x161; = dx, 2 y = x/2 + C, 2 = C, y 2 â&#x2C6;&#x161; y = x/4 + 1, y = (x/4 + 1)2 ,

(c)

y(1) = 25/16 = 1.5625

dy = ky 2 , y(0) = y0 , k > 0 dt

(b)

3. (a)

ds 1 = s dt 2

(b)

4. (a)

dv = â&#x2C6;&#x2019;2v 2 dt

dy = â&#x2C6;&#x2019;ky 2 , y(0) = y0 , k > 0 dt

1. (a)

EXERCISE SET 9.3

dy = 0.01y, y0 = 10,000 dt 1 1 ln 2 â&#x2030;&#x2C6; 69.31 h (c) T = ln 2 = k 0.01

1 1 ln 2 = ln 2 20 T dy (a) = ((ln 2)/20)y, y(0) = 1 dt

6. k =

ds d2 s =2 dt2 dt d2 s = â&#x2C6;&#x2019;2 dt2

an (b)

5. (a)

ds dt

(b) y = 10,000et/100 (d) 45,000 = 10,000et/100 , 45,000 â&#x2030;&#x2C6; 150.41 h t = 100 ln 10,000

(b) y(t) = et(ln 2)/20 = 2t/20 (d) 1,000,000 = 2t/20 , t = 20

ln 106 â&#x2030;&#x2C6; 398.63 min ln 2

uh a

1 ln 2 dy = â&#x2C6;&#x2019;ky, y(0) = 5.0 Ă&#x2014; 107 ; 3.83 = T = ln 2, so k = â&#x2030;&#x2C6; 0.1810 dt k 3.83 (b) y = 5.0 Ă&#x2014; 107 eâ&#x2C6;&#x2019;0.181t

7. (a)

(c) y(30) = 5.0 Ă&#x2014; 107 eâ&#x2C6;&#x2019;0.1810(30) â&#x2030;&#x2C6; 219,000 (d) y(t) = (0.1)y0 = y0 eâ&#x2C6;&#x2019;kt , â&#x2C6;&#x2019;kt = ln 0.1, t = â&#x2C6;&#x2019;

ln 0.1 = 12.72 days 0.1810

1 1 dy ln 2 = ln 2 â&#x2030;&#x2C6; 0.0050, so = â&#x2C6;&#x2019;0.0050y, y0 = 10. T 140 dt (b) y = 10eâ&#x2C6;&#x2019;0.0050t

8. (a) k =

3/2.

â&#x2C6;&#x161; 20. (a) y0 = 1, yn+1 = yn + ( yn /2)â&#x2C6;&#x2020;x â&#x2C6;&#x161; â&#x2C6;&#x2020;x = 0.2 : yn+1 = yn + yn /10; y5 â&#x2030;&#x2C6; 1.5489 â&#x2C6;&#x161; â&#x2C6;&#x2020;x = 0.1 : yn+1 = yn + yn /20; y10 â&#x2030;&#x2C6; 1.5556 â&#x2C6;&#x161; â&#x2C6;&#x2020;x = 0.05 : yn+1 = yn + yn /40; y20 â&#x2030;&#x2C6; 1.5590

â&#x2C6;&#x161;

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Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Exercise Set 9.3

us uf

9. 100e0.02t = 5000, e0.02t = 50, t =

383

1 ln 50 ≈ 196 days 0.02

1 10. y = 10,000ekt , but y = 12,000 when t = 10 so 10,000e10k = 12,000, k = ln 1.2. y = 20,000 when 10 ln 2 ln 2 = 10 ≈ 38, in the year 2025. 2 = ekt , t = k ln 1.2 3.5 1 1 ≈ 0.2100, T = ln 2 ≈ 3.30 days 11. y(t) = y0 e−kt = 10.0e−kt , 3.5 = 10.0e−k(5) , k = − ln 5 10.0 k

1 12. y = y0 e−kt , 0.6y0 = y0 e−5k , k = − ln 0.6 ≈ 0.10 5 ln 2 (a) T = ≈ 6.9 yr k y (b) y(t) ≈ y0 e−0.10t , ≈ e−0.10t , so e−0.10t × 100 percent will remain. y0 ln 2 ≈ 0.1386; y ≈ 2e0.1386t 5

13. (a) k =

(b) y(t) = 5e0.015t 1 ln 100 ≈ 0.5117, 9 ≈ 0.5995, y ≈ 0.5995e0.5117t .

y ≈ y0 e0.5117t ; also y(1) = 1, so y0 = e−0.5117

ln 2 ≈ 0.1386, 1 = y(1) ≈ y0 e0.1386 , y0 ≈ e−0.1386 ≈ 0.8706, y ≈ 0.8706e0.1386t T

(d) k =

ln 2 ≈ 0.1386, y ≈ 10e−0.1386t T

14. (a) k =

(b) y = 10e−0.015t

1 ln 100 ≈ 0.5117; 9

ln 2 ≈ 0.1386, 10 = y(1) ≈ y0 e−0.1386 , y0 ≈ 10e0.1386 ≈ 11.4866, y ≈ 11.4866e−0.1386t T

(d) k =

16. (a) None; the half-life is independent of the initial amount. (b) kT = ln 2, so T is inversely proportional to k. ln 2 ; and ln 2 ≈ 0.6931. If k is measured in percent, k = 100k, k 69.31 70 ln 2 ≈ ≈ . then T = k k k

17. (a) T =

(b) 70 yr

(d) 7%

uh a

18. Let y = y0 ekt with y = y1 when t = t1 and y = 3y1 when t = t1 + T ; then y0 ekt1 = y1 (i) and 1 y0 ek(t1 +T ) = 3y1 (ii). Divide (ii) by (i) to get ekT = 3, T = ln 3. k

19. From (12), y(t) = y0 e−0.000121t . If 0.27 = if 0.30 =

y(t) ln 0.27 = e−0.000121t then t = − ≈ 10,820 yr, and y0 0.000121

ln 0.30 y(t) ≈ 9950, or roughly between 9000 B.C. and 8000 B.C. then t = − y0 0.000121

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Chapter 9

20. (a)

(b) t = 1988 yields

us uf

y/y0 = eâ&#x2C6;&#x2019;0.000121(1988) â&#x2030;&#x2C6; 79%.

50000 0

6eâ&#x2C6;&#x2019;2k = 2, k =

2Âˇ8 16 passes through the point (2, 4), 4 = , â&#x2C6;&#x2019;kt 2 + 6e 2 + 6eâ&#x2C6;&#x2019;2k

21. y0 â&#x2030;&#x2C6; 2, L â&#x2030;&#x2C6; 8; since the curve y = 1 ln 3 â&#x2030;&#x2C6; 0.5493. 2

400,000 passes through the point (200, 600), 400 + 600eâ&#x2C6;&#x2019;kt 800 1 = , k= ln 2.25 â&#x2030;&#x2C6; 0.00405. 3 200

400,000 , 600eâ&#x2C6;&#x2019;200k 400 + 600eâ&#x2C6;&#x2019;200k

23. (a) y0 = 5

(b) L = 12

(d) L/2 = 6 =

60 , 5 + 7eâ&#x2C6;&#x2019;t = 10, t = â&#x2C6;&#x2019; ln(5/7) â&#x2030;&#x2C6; 0.3365 5 + 7eâ&#x2C6;&#x2019;t

dy 1 = y(12 â&#x2C6;&#x2019; y), y(0) = 5 dt 12

1 1000 , 3(1 + 999eâ&#x2C6;&#x2019;0.9t ) = 4, t = ln(3 Âˇ 999) â&#x2030;&#x2C6; 8.8949 â&#x2C6;&#x2019;0.9t 1 + 999e 0.9

dy 0.9 = y(1000 â&#x2C6;&#x2019; y), y(0) = 1 dt 1000

25. See (13):

dy = 10(1 â&#x2C6;&#x2019; 0.1y)y = 25 â&#x2C6;&#x2019; (y â&#x2C6;&#x2019; 5)2 is maximized when y = 5. dt

dy 1 = 50y 1 â&#x2C6;&#x2019; y ; from (13), k = 50, L = 50,000. dt 50,000

26.

(b) k = 10

(a) L = 10 (c)

(e)

(b) L = 1000

24. (a) y0 = 1 (d) 750 =

(e)

600 =

22. y0 â&#x2030;&#x2C6; 400, L â&#x2030;&#x2C6; 1000; since the curve y =

(a) L = 50,000 d dy is maximized when 0 = dt dy

uh a

(c)

(b) k = 50

dy dt

= 50 â&#x2C6;&#x2019; y/500, y = 25,000

27. Assume y(t) students have had the ďŹ&#x201A;u t days after semester break. Then y(0) = 20, y(5) = 35. (a)

dy = ky(L â&#x2C6;&#x2019; y) = ky(1000 â&#x2C6;&#x2019; y), y0 = 20 dt

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Exercise Set 9.3

20000 1000 = ; 20 + 980eâ&#x2C6;&#x2019;kt 1 + 49eâ&#x2C6;&#x2019;kt 1000 1000 35 = , k = 0.115, y â&#x2030;&#x2C6; . 1 + 49eâ&#x2C6;&#x2019;0.115t 1 + 49eâ&#x2C6;&#x2019;5k

385

t 0 1 2 3 y(t) 20 22 25 28

us uf

(c)

(b) Part (a) has solution y =

4 5 6 7 8 9 10 11 12 13 14 31 35 39 44 49 54 61 67 75 83 93

(d) 100

75 50 25 t 6

dp = â&#x2C6;&#x2019;kp, p(0) = p0 dh (b) p0 = 1, so p = eâ&#x2C6;&#x2019;kh , but p = 0.83 when h = 5000 thus eâ&#x2C6;&#x2019;5000k = 0.83, k=â&#x2C6;&#x2019;

ln 0.83 â&#x2030;&#x2C6; 0.0000373, p â&#x2030;&#x2C6; eâ&#x2C6;&#x2019;0.0000373h atm. 5000

dT dT = â&#x2C6;&#x2019;k(T â&#x2C6;&#x2019; 21), T (0) = 95, = â&#x2C6;&#x2019;k dt, ln(T â&#x2C6;&#x2019; 21) = â&#x2C6;&#x2019;kt + C1 , dt T â&#x2C6;&#x2019; 21

29. (a)

28. (a)

T = 21 + eC1 eâ&#x2C6;&#x2019;kt = 21 + Ceâ&#x2C6;&#x2019;kt , 95 = T (0) = 21 + C, C = 74, T = 21 + 74eâ&#x2C6;&#x2019;kt

32 37

t ,

dT = k(70 â&#x2C6;&#x2019; T ), T (0) = 40; â&#x2C6;&#x2019; ln(70 â&#x2C6;&#x2019; T ) = kt + C, 70 â&#x2C6;&#x2019; T = eâ&#x2C6;&#x2019;kt eâ&#x2C6;&#x2019;C , T = 40 when t = 0, so dt 70 â&#x2C6;&#x2019; 52 5 30 = eâ&#x2C6;&#x2019;C , T = 70 â&#x2C6;&#x2019; 30eâ&#x2C6;&#x2019;kt ; 52 = T (1) = 70 â&#x2C6;&#x2019; 30eâ&#x2C6;&#x2019;k , k = â&#x2C6;&#x2019; ln = ln â&#x2030;&#x2C6; 0.5, 30 3 T â&#x2030;&#x2C6; 70 â&#x2C6;&#x2019; 30eâ&#x2C6;&#x2019;0.5t

30.

32 64 = â&#x2C6;&#x2019; ln , T = 21 + 74et ln(32/37) = 21 + 74 (b) 85 = T (1) = 21 + 74e , k = â&#x2C6;&#x2019; ln 74 37 t 32 ln(30/74) 30 = , t= T = 51 when â&#x2030;&#x2C6; 6.22 min 37 ln(32/37) 74 â&#x2C6;&#x2019;k

uh a

31. Let T denote the body temperature of McHamâ&#x20AC;&#x2122;s body at time t, the number of hours elapsed after dT dT = â&#x2C6;&#x2019;kdt, ln(T â&#x2C6;&#x2019; 72) = â&#x2C6;&#x2019;kt + C, T = 72 + eC eâ&#x2C6;&#x2019;kt , 10:06 P.M.; then = â&#x2C6;&#x2019;k(T â&#x2C6;&#x2019; 72), dt T â&#x2C6;&#x2019; 72 3.6 77.9 = 72 + eC , eC = 5.9, T = 72 + 5.9eâ&#x2C6;&#x2019;kt , 75.6 = 72 + 5.9eâ&#x2C6;&#x2019;k , k = â&#x2C6;&#x2019; ln â&#x2030;&#x2C6; 0.4940, 5.9 ln(26.6/5.9) â&#x2030;&#x2C6; â&#x2C6;&#x2019;3.05, T = 72+5.9eâ&#x2C6;&#x2019;0.4940t . McHamâ&#x20AC;&#x2122;s body temperature was last 98.6â&#x2014;Ś when t = â&#x2C6;&#x2019; 0.4940 so around 3 hours and 3 minutes before 10:06; the death took place at approximately 7:03 P.M., while Moore was on stage. dT dT = k(Ta â&#x2C6;&#x2019; T ) where k > 0. If T0 > Ta then = â&#x2C6;&#x2019;k(T â&#x2C6;&#x2019; Ta ) where k > 0; dt dt â&#x2C6;&#x2019;kt with k > 0. both cases yield T (t) = Ta + (T0 â&#x2C6;&#x2019; Ta )e

32. If T0 < Ta then

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Chapter 9

33. (a) y = y0 bt = y0 et ln b = y0 ekt with k = ln b > 0 since b > 1.

34. If y = y0 ekt and y = y1 = y0 ekt1 then y1 /y0 = ekt1 , k =

ln(y1 /y0 ) ; if y = y0 e−kt and t1

ln(y1 /y0 ) . t1

y = y1 = y0 e−kt1 then y1 /y0 = e−kt1 , k = −

us uf

(d) y = 4(0.5t ) = 4et ln 0.5 = 4e−t ln 2

(b) y = y0 bt = y0 et ln b = y0 e−kt with k = − ln b > 0 since 0 < b < 1.

EXERCISE SET 9.4

1. (a) y = e2x , y = 2e2x , y = 4e2x ; y − y − 2y = 0 y = e−x , y = −e−x , y = e−x ; y − y − 2y = 0.

(b) y = c1 e2x + c2 e−x , y = 2c1 e2x − c2 e−x , y = 4c1 e2x + c2 e−x ; y − y − 2y = 0

2. (a) y = e−2x , y = −2e−2x , y = 4e−2x ; y + 4y + 4y = 0 y = xe−2x , y = (1 − 2x)e−2x , y = (4x − 4)e−2x ; y + 4y + 4y = 0. (b) y = c1 e−2x + c2 xe−2x , y = −2c1 e−2x + c2 (1 − 2x)e−2x , y = 4c1 e−2x + c2 (4x − 4)e−2x ; y + 4y + 4y = 0.

3. m2 + 3m − 4 = 0, (m − 1)(m + 4) = 0; m = 1, −4 so y = c1 ex + c2 e−4x . 4. m2 + 6m + 5 = 0, (m + 1)(m + 5) = 0; m = −1, −5 so y = c1 e−x + c2 e−5x .

5. m2 − 2m + 1 = 0, (m − 1)2 = 0; m = 1, so y = c1 ex + c2 xex .

6. m2 + 6m + 9 = 0, (m + 3)2 = 0; m = −3 so y = c1 e−3x + c2 xe−3x . √ √ √ 7. m2 + 5 = 0, m = ± 5 i so y = c1 cos 5 x + c2 sin 5 x. 8. m2 + 1 = 0, m = ±i so y = c1 cos x + c2 sin x.

9. m2 − m = 0, m(m − 1) = 0; m = 0, 1 so y = c1 + c2 ex . 10. m2 + 3m = 0, m(m + 3) = 0; m = 0, −3 so y = c1 + c2 e−3x .

11. m2 + 4m + 4 = 0, (m + 2)2 = 0; m = −2 so y = c1 e−2t + c2 te−2t .

12. m2 − 10m + 25 = 0, (m − 5)2 = 0; m = 5 so y = c1 e5t + c2 te5t . 13. m2 − 4m + 13 = 0, m = 2 ± 3i so y = e2x (c1 cos 3x + c2 sin 3x).

uh a

14. m2 − 6m + 25 = 0, m = 3 ± 4i so y = e3x (c1 cos 4x + c2 sin 4x). 15. 8m2 − 2m − 1 = 0, (4m + 1)(2m − 1) = 0; m = −1/4, 1/2 so y = c1 e−x/4 + c2 ex/2 . 16. 9m2 − 6m + 1 = 0, (3m − 1)2 = 0; m = 1/3 so y = c1 ex/3 + c2 xex/3 .

17. m2 + 2m − 3 = 0, (m + 3)(m − 1) = 0; m = −3, 1 so y = c1 e−3x + c2 ex and y = −3c1 e−3x + c2 ex . Solve the system c1 + c2 = 1, −3c1 + c2 = 5 to get c1 = −1, c2 = 2 so y = −e−3x + 2ex .

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Exercise Set 9.4

387

18. m2 â&#x2C6;&#x2019; 6m â&#x2C6;&#x2019; 7 = 0, (m + 1)(m â&#x2C6;&#x2019; 7) = 0; m = â&#x2C6;&#x2019;1, 7 so y = c1 eâ&#x2C6;&#x2019;x + c2 e7x , y = â&#x2C6;&#x2019;c1 eâ&#x2C6;&#x2019;x + 7c2 e7x . Solve the system c1 + c2 = 5, â&#x2C6;&#x2019;c1 + 7c2 = 3 to get c1 = 4, c2 = 1 so y = 4eâ&#x2C6;&#x2019;x + e7x .

us uf

19. m2 â&#x2C6;&#x2019; 6m + 9 = 0, (m â&#x2C6;&#x2019; 3)2 = 0; m = 3 so y = (c1 + c2 x)e3x and y = (3c1 + c2 + 3c2 x)e3x . Solve the system c1 = 2, 3c1 + c2 = 1 to get c1 = 2, c2 = â&#x2C6;&#x2019;5 so y = (2 â&#x2C6;&#x2019; 5x)e3x .

â&#x2C6;&#x161; â&#x2C6;&#x161; â&#x2C6;&#x161; 1 = 0, m =â&#x2C6;&#x161;â&#x2C6;&#x2019;2 Âą 3 so yâ&#x2C6;&#x161;= c1 e(â&#x2C6;&#x2019;2+ â&#x2C6;&#x161;3)x + c2 e(â&#x2C6;&#x2019;2â&#x2C6;&#x2019; 3)x , 20. m2 + 4m + â&#x2C6;&#x161; (â&#x2C6;&#x2019;2+ 3)x + (â&#x2C6;&#x2019;2 â&#x2C6;&#x2019; 3)c2 e(â&#x2C6;&#x2019;2â&#x2C6;&#x2019; 3)x c1 + c2 = 5, y = (â&#x2C6;&#x2019;2 â&#x2C6;&#x161; . Solve the system â&#x2C6;&#x161; â&#x2C6;&#x161; + 3)c1 e â&#x2C6;&#x161; (â&#x2C6;&#x2019;2 + 3)c1 + (â&#x2C6;&#x2019;2 â&#x2C6;&#x2019; 3)c2 = 4 to get c1 = 52 + 73 3, c2 = 52 â&#x2C6;&#x2019; 73 3 so â&#x2C6;&#x161; â&#x2C6;&#x161; â&#x2C6;&#x161; â&#x2C6;&#x161; y = 52 + 37 3 e(â&#x2C6;&#x2019;2+ 3)x + 52 â&#x2C6;&#x2019; 73 3 e(â&#x2C6;&#x2019;2â&#x2C6;&#x2019; 3)x .

21. m2 + 4m + 5 = 0, m = â&#x2C6;&#x2019;2 Âą i so y = eâ&#x2C6;&#x2019;2x (c1 cos x + c2 sin x), y = eâ&#x2C6;&#x2019;2x [(c2 â&#x2C6;&#x2019; 2c1 ) cos x â&#x2C6;&#x2019; (c1 + 2c2 ) sin x]. Solve the system c1 = â&#x2C6;&#x2019;3, c2 â&#x2C6;&#x2019; 2c1 = 0 to get c1 = â&#x2C6;&#x2019;3, c2 = â&#x2C6;&#x2019;6 so y = â&#x2C6;&#x2019;eâ&#x2C6;&#x2019;2x (3 cos x + 6 sin x).

22. m2 â&#x2C6;&#x2019; 6m + 13 = 0, m = 3 Âą 2i so y = e3x (c1 cos 2x + c2 sin 2x), y = e3x [(3c1 + 2c2 ) cos 2x â&#x2C6;&#x2019; (2c1 â&#x2C6;&#x2019; 3c2 ) sin 2x]. Solve the system c1 = â&#x2C6;&#x2019;1, 3c1 + 2c2 = 1 to get c1 = â&#x2C6;&#x2019;1, c2 = 2 so y = e3x (â&#x2C6;&#x2019; cos 2x + 2 sin 2x).

23. (a) m = 5, â&#x2C6;&#x2019;2 so (m â&#x2C6;&#x2019; 5)(m + 2) = 0, m2 â&#x2C6;&#x2019; 3m â&#x2C6;&#x2019; 10 = 0; y â&#x2C6;&#x2019; 3y â&#x2C6;&#x2019; 10y = 0. (b) m = 4, 4 so (m â&#x2C6;&#x2019; 4)2 = 0, m2 â&#x2C6;&#x2019; 8m + 16 = 0; y â&#x2C6;&#x2019; 8y + 16y = 0. (c) m = â&#x2C6;&#x2019;1 Âą 4i so (m + 1 â&#x2C6;&#x2019; 4i)(m + 1 + 4i) = 0, m2 + 2m + 17 = 0; y + 2y + 17y = 0.

24. c1 ex + c2 eâ&#x2C6;&#x2019;x is the general solution, but cosh x = 12 ex + 12 eâ&#x2C6;&#x2019;x and sinh x = 12 ex â&#x2C6;&#x2019; 12 eâ&#x2C6;&#x2019;x so cosh x and sinh x are also solutions.

â&#x2C6;&#x161; 25. m2 + km + k = 0, m = â&#x2C6;&#x2019;k Âą k 2 â&#x2C6;&#x2019; 4k /2 (a) k 2 â&#x2C6;&#x2019; 4k > 0, k(k â&#x2C6;&#x2019; 4) > 0; k < 0 or k > 4 (b) k 2 â&#x2C6;&#x2019; 4k = 0; k = 0, 4

dy dz 1 dy dy = = and dx dz dx x dz d 1 dy 1 d2 y dz d2 y d dy 1 dy 1 dy 1 d2 y = = = â&#x2C6;&#x2019; 2 â&#x2C6;&#x2019; 2 = 2 , 2 2 2 dx dx dx dx x dz x dz dx x dz x dz x dz dy d2 y substitute into the original equation to get 2 + (p â&#x2C6;&#x2019; 1) + qy = 0. dz dz

26. z = ln x;

dy d2 y +2 + 2y = 0, m2 + 2m + 2 = 0; m = â&#x2C6;&#x2019;1 Âą i so 2 dz dz 1 y = eâ&#x2C6;&#x2019;z (c1 cos z + c2 sin z) = [c1 cos(ln x) + c2 sin(ln x)]. x â&#x2C6;&#x161; d2 y dy (b) â&#x2C6;&#x2019;2 â&#x2C6;&#x2019; 2y = 0, m2 â&#x2C6;&#x2019; 2m â&#x2C6;&#x2019; 2 = 0; m = 1 Âą 3 so 2 dz dzâ&#x2C6;&#x161; â&#x2C6;&#x161; â&#x2C6;&#x161; â&#x2C6;&#x161; y = c1 e(1+ 3)z + c2 e(1â&#x2C6;&#x2019; 3)z = c1 x1+ 3 + c2 x1â&#x2C6;&#x2019; 3

27. (a)

uh a

28. m2 + pm + q = 0, m = 12 (â&#x2C6;&#x2019;p Âą p2 â&#x2C6;&#x2019; 4q ). If 0 < q < p2 /4 then y = c1 em1 x + c2 em2 x where â&#x2C6;&#x2019;px/2 m1 < 0 and m2 < 0, if q = p2 /4 then y = c1 e + c2 xeâ&#x2C6;&#x2019;px/2 , if q > p2 /4 then 1 â&#x2C6;&#x2019;px/2 (c1 cos kx + c2 sin kx) where k = 2 4q â&#x2C6;&#x2019; p2 . In all cases lim y(x) = 0. y=e xâ&#x2020;&#x2019;+â&#x2C6;&#x17E;

29. (a) Neither is a constant multiple of the other, since, e.g. if y1 = ky2 then em1 x = kem2 x , e(m1 â&#x2C6;&#x2019;m2 )x = k. But the right hand side is constant, and the left hand side is constant only if m1 = m2 , which is false.

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Chapter 9

(b) If y1 = ky2 then emx = kxemx , kx = 1 which is impossible. If y2 = y1 then xemx = kemx , x = k which is impossible.

us uf

30. y1 = eax cos bx, y1 = eax (a cos bx − b sin bx), y1 = eax [(a2 − b2 ) cos bx − 2ab sin bx] so y1 + py1 + qy1 = eax [(a2 − b2 + ap + q) cos bx − (2ab + bp) sin bx]. But a = − 12 p and b = 12 4q − p2 so a2 − b2 + ap + q = 0 and 2ab + bp = 0 thus y1 + py1 + qy1 = 0. Similarly, y2 = eax sin bx is also a solution. Since y1 /y2 = cot bx and y2 /y1 = tan bx it is clear that the two solutions are linearly independent.

(b)

31. (a) The general solution is c1 eµx + c2 emx ; let c1 = 1/(µ − m), c2 = −1/(µ − m). eµx − emx = lim xeµx = xemx . µ→m µ→m µ−m lim

32. (a) If λ = 0, then y = 0, y = c1 + c2 x. Use y(0) = 0 and y(π) = 0 to get c1 = c2 = 0. If λ < 0, then let λ = −a2 where a > 0 so y − a2 y = 0, y = c1 eax + c2 e−ax . Use y(0) = 0 and y(π) = 0 to get c1 = c2 = 0. √ √ √ (b) If λ > 0, then m2 + λ = 0, m2 = −λ = λi2 , m√= ± λi, y √ = c1 cos λ x + c2√sin λ x. If y(0) = √ 0 and y(π) = 0, then c1 = 0 and c1 cos π √ λ + c2 sin √ π λ = 0 so c2 sin π λ = 0. But c2 sin π λ = 0 for arbitrary values of c2 if sin π λ = 0, π λ = nπ, λ = n2 for n = 1, 2, 3, . . ., otherwise c2 = 0. 33. k/M = 0.25/1 = 0.25

(d) y = 0 at the equilibrium position, so t/2 = π/2, t = π s.

0.3

(c)

(b) T = 2π · 2 = 4π s, f = 1/T = 1/(4π) Hz

(a) From (20), y = 0.3 cos(t/2)

x 4c

-0.3

t/2 = π at the maximum position below the equlibrium position, so t = 2π s.

√ k/M = 1/(2 2)

(c)

34. 64 = w = −M g, M = 2, k/M = 0.25/2 = 1/8, √ (a) From (20), y = cos(t/(2 2)) √ √ M = 2π(2 2) = 4π 2 s, (b) T = 2π k √ f = 1/T = 1/(4π 2) Hz

(e)

t 6p

10p

uh a

-1

√ √ (d) y = 0 at the equilibrium position, so t/(2 2) = π/2, t = π 2 s √ √ (e) t/(2 2) = π, t = 2π 2 s

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35. l = 0.05, k/M = g/l = 9.8/0.05 = 196 s−2

(c)

(d) 14t = π/2, t = π/28 s

0.15

(e)

14t = π, t = π/14 s

t π 7

2π 7

-0.15

k/M = 8

(b) T = 2π M/k = 2π/8 = π/4 s; f = 1/T = 4/π Hz

(a) From (20), y = −1.5 cos 8t. (c)

36. l = 0.5, k/M = g/l = 32/0.5 = 64,

(d) 8t = π/2, t = π/16 s

t 6

dy k t, so v = = −y0 M dt

37. Assume y = y0 cos

-2

(a) The maximum speed occurs when sin k t = 0, y = 0. so cos M (b) The minimum speed occurs when sin

38. (a) T = 2π

25w = 9w + 36, w =

k sin M

k t M

k t = ±1, M

k t = 0, M

k t = nπ + π/2, M

k t = nπ, so cos M

k t = ±1, y = ±y0 . M

M 4π 2 w 4π 2 4π 2 w 4π 2 w + 4 , k = 2 M = 2 , so k = = , 25w = 9(w + 4), k T T g g 9 g 25

8t = π, t = π/8 s

(e)

(b) From Part (a), w =

uh a

us uf

(b) T = 2π M/k = 2π/14 = π/7 s, f = 7/π Hz

(a) From (20), y = −0.12 cos 14t.

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Exercise Set 9.4

4π 2 w 4π 2 1 π2 9 ,k = = = 4 g 9 32 4 32 9 4

39. By Hooke’s Law, F (t) = −kx(t), since the only force is the restoring force of the spring. Newton’s Second Law gives F (t) = M x (t), so M x (t) + kx(t) = 0, x(0) = x0 , x (0) = 0. k k , so c2 = 0; y0 = y(0) = c1 , so y = y0 cos t. 40. 0 = v(0) = y (0) = c2 M M

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Chapter 9

41. (a) m2 + 2.4m + 1.44 = 0, (m + 1.2)2 = 0, m = −1.2, y = C1 e−6t/5 + C2 te−6t/5 ,

6 16 16 , y = e−6t/5 + te−6t/5 C1 = 1, 2 = y (0) = − C1 + C2 , C2 = 5 5 5

us uf

y 2

(b) y (t) = 0 when t = t1 = 25/48 ≈ 0.520833, y(t1 ) = 1.427364 cm 16 −6t/5 (t + 5/16) = 0 only if t = −5/16, so y = 0 for t ≥ 0. e 5 √ 42. (a) m2 + 5m + 2 = (m + 5/2)2 − 17/4 = 0, m = −5/2 ± 17/2, y = C1 e(−5+

√ 17)t/2

+ C2 e(−5−

, √

x 1

√ −5 + 17 −5 − 17 C1 + C2 = 1/2, −4 = y (0) = C1 + C2 2 2 √ √ 17 − 11 17 17 + 11 17 , C2 = C1 = 68 68 √ √ √ 17 − 11 17 (−5+ 17)t/2 17 + 11 17 −(5+√17)t/2 e y= + e 68 68

0.5 0.4

0.2 0.1

0.3

-0.1

-0.2

-0.3 -0.4

-0.5

(b) y (t) = 0 when t = t1 = 0.759194, y(t1 ) = −0.270183 cm so the maximum distance below the equilibrium position is 0.270183 cm.

uh a

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us uf

√ √ √ 43. (a) m2 + m + 5 = 0, m = −1/2 ± ( 19/2)i, y = e−t/2 C1 cos( 19t/2) + C2 sin( 19t/2) , √ √ 1 = y(0) = C1 , −3.5 = y (0) = −(1/2)C1 + ( 19/2)C2 , C2 = −6/ 19, √ √ √ y = e−t/2 cos( 19 t/2) − (6/ 19 )e−t/2 sin( 19 t/2) y 1 x 2

391

Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Exercise Set 9.4

-1

(b) y (t) = 0 for the ﬁrst time when t = t1 = 0.905533, y(t1 ) = −1.054466 cm so the maximum distance below the equilibrium position is 1.054466 cm.

(c) y(t) = 0 for the ﬁrst time when t = t2 = 0.288274, y (t2 ) = −3.210357 cm/s. (d) The acceleration is y (t) so from the diﬀerential equation y = −y − 5y. But y = 0 when the object passes through the equilibrium position, thus y = −y = 3.210357 cm/s2 . √ √ 11i/2, y = e−t/2 C1 cos( 11t/2) + C2 sin( 11t/2) , √ √ −2 = y(0) = C1 , v0 = y (0) = −(1/2)C1 + ( 11/2)C2 , C2 = (v0 − 1)(2/ 11), √ √ √ y(t) = e−t/2 −2 cos( 11 t/2) + (2/ 11 )(v0 − 1) sin( 11 t/2) √ √ √ y (t) = e−t/2 v0 cos( 11t/2) + (12 − v0 )/ 11 sin( 11t/2)

√

44. (a) m2 + m + 3m = 0, m = −1/2 ±

uh a

(b) We wish to ﬁnd v0 such that y(t) = 1 but no greater. This implies that y (t) = 0 at that point. So ﬁnd the largest √ value of v0 such √ that there is a solution of y (t) = 0, y(t) = 1. Note 11 v0 11 that y (t) = 0 when tan . Choose the smallest positive solution t0 of this t= 2 v0 − 12 equation. Then √ √ 11 11 12[(v0 − 1)2 + 11] 2 2 t0 = 1 + tan t0 = . sec 2 2 (v0 − 12)2 √ 11 t0 < 0, so Assume for now that v0 < 12; if not, we will deal with it later. Then tan 2 √ √ √ 2 3 (v0 − 1)2 + 11 π 11 11 t0 < π, and sec t0 = < 2 2 v0 − 12 2 √ 11 v0 − 12 t0 = √ and cos , 2 3 (v0 − 1)2 + 11 √ √ √ √ 11 11 11 v0 11 t0 = tan t0 cos t0 = √ sin , and 2 2 2 2 3 (v0 − 1)2 + 11 √ √ (v0 − 1)2 + 11 11 − 1) 11 2(v 0 −t0 /2 −t0 /2 √ √ y(t0 ) = e −2 cos sin . t0 + t0 = e 2 2 11 3

Use various values of v0 , 0 < v0 < 12 to determine the transition point from y < 1 to y > 1 and then reﬁne the partition on the values of v to arrive at v ≈ 2.44 cm/s.

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392

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Chapter 9

(c)

t 6

us uf

45. (a) m2 + 3.5m + 3 = (m + 1.5)(m + 2), y = C1 eâ&#x2C6;&#x2019;3t/2 + C2 eâ&#x2C6;&#x2019;2t , y(t) = (4 + 2v0 )eâ&#x2C6;&#x2019;3t/2 â&#x2C6;&#x2019; (3 + 2v0 )eâ&#x2C6;&#x2019;2t (b) v0 = 2, y(t) = 8eâ&#x2C6;&#x2019;3t/2 â&#x2C6;&#x2019; 7eâ&#x2C6;&#x2019;2t , v0 = â&#x2C6;&#x2019;1, y(t) = 2eâ&#x2C6;&#x2019;3t/2 â&#x2C6;&#x2019; eâ&#x2C6;&#x2019;2t , v0 = â&#x2C6;&#x2019;4, y(t) = â&#x2C6;&#x2019;4eâ&#x2C6;&#x2019;3t/2 + 5eâ&#x2C6;&#x2019;2t

y 2

v0 = â&#x20AC;&#x201C;1 1

v0 = â&#x20AC;&#x201C;4

-1

dy1 dy dy1 + p(x)y = c + p(x)(cy1 ) = c + p(x)y1 = c Âˇ 0 = 0 dt dt dt

46.

v0 = 2

1 = y(0) = C1 + C2 , v0 = y (0) = â&#x2C6;&#x2019;(3/2)C1 â&#x2C6;&#x2019; 2C2 , C1 = 4 + 2v0 , C2 = â&#x2C6;&#x2019;3 â&#x2C6;&#x2019; 2v0 ,

CHAPTER 9 SUPPLEMENTARY EXERCISES 4. The diďŹ&#x20AC;erential equation in Part (c) is not separable; the others are. linear

(b) 2

6. IF: Âľ = eâ&#x2C6;&#x2019;2x ,

linear and separable

(c)

separable

(d)

neither

2 2 2 2 d â&#x2C6;&#x2019;2x2 1 1 ye = xeâ&#x2C6;&#x2019;2x , yeâ&#x2C6;&#x2019;2x = â&#x2C6;&#x2019; eâ&#x2C6;&#x2019;2x + C, y = â&#x2C6;&#x2019; + Ce2x dx 4 4

5. (a)

2 2 2 1 x2 dy 1 = x dx, ln |4y +1| = +C1 , 4y +1 = Âąe4C1 e2x = C2 e2x ; y = â&#x2C6;&#x2019; +Ce2x , 4y + 1 4 2 4 including C = 0

Sep of var:

uh a

7. The parabola ky(L â&#x2C6;&#x2019; y) opens down and has its maximum midway between the y-intercepts, that dy 1 is, at the point y = (0 + L) = L/2, where = k(L/2)2 = kL2 /4. dt 2

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Chapter 9 Supplementary Exercises

(t2 â&#x2C6;&#x2019; t1 ) ln 2 ln 2 = . If y = y0 eâ&#x2C6;&#x2019;kt , then y1 = y0 eâ&#x2C6;&#x2019;kt1 , y2 = y0 eâ&#x2C6;&#x2019;kt2 , k ln(y2 /y1 )

y2 /y1 = eâ&#x2C6;&#x2019;k(t2 â&#x2C6;&#x2019;t1 ) , k = â&#x2C6;&#x2019;

1 ln 2 (t2 â&#x2C6;&#x2019; t1 ) ln 2 ln(y2 /y1 ), T = =â&#x2C6;&#x2019; . t2 â&#x2C6;&#x2019; t1 k ln(y2 /y1 )

(t2 â&#x2C6;&#x2019; t1 ) ln 2

. In either case, T is positive, so T =

ln(y2 /y1 )

ln 2 â&#x2030;&#x2C6; 3.1 h. ln 1.25

(b) In Part (a) assume t2 = t1 + 1 and y2 = 1.25y1 . Then T =

T =

1 ln(y2 /y1 ), t2 â&#x2C6;&#x2019; t1

us uf

8. (a) If y = y0 ekt , then y1 = y0 ekt1 , y2 = y0 ekt2 , divide: y2 /y1 = ek(t2 â&#x2C6;&#x2019;t1 ) , k =

393

4Ď&#x20AC; 3 dV dr dV = â&#x2C6;&#x2019;kS; but V = r , = 4Ď&#x20AC;r2 , S = 4Ď&#x20AC;r2 , so dr/dt = â&#x2C6;&#x2019;k, r = â&#x2C6;&#x2019;kt + C, 4 = C, dt dt dt 3

r = â&#x2C6;&#x2019;kt + 4, 3 = â&#x2C6;&#x2019;k + 4, k = 1, r = 4 â&#x2C6;&#x2019; t m.

10. Assume the tank contains y(t) oz of salt at time t. Then y0 = 0 and for 0 < t < 15, dy y = 5 Âˇ 10 â&#x2C6;&#x2019; 10 = (50 â&#x2C6;&#x2019; y/100) oz/min, with solution y = 5000 + Ceâ&#x2C6;&#x2019;t/100 . But y(0) = 0 so dt 1000 C = â&#x2C6;&#x2019;5000, y = 5000(1 â&#x2C6;&#x2019; eâ&#x2C6;&#x2019;t/100 ) for 0 â&#x2030;¤ t â&#x2030;¤ 15, and y(15) = 5000(1 â&#x2C6;&#x2019; eâ&#x2C6;&#x2019;0.15 ). For 15 < t < 30, dy y = 0â&#x2C6;&#x2019; 5, y = C1 eâ&#x2C6;&#x2019;t/200 , C1 eâ&#x2C6;&#x2019;0.075 = y(15) = 5000(1â&#x2C6;&#x2019;eâ&#x2C6;&#x2019;0.15 ), C1 = 5000(e0.075 â&#x2C6;&#x2019;eâ&#x2C6;&#x2019;0.075 ), dt 1000 y = 5000(e0.075 â&#x2C6;&#x2019; eâ&#x2C6;&#x2019;0.075 )eâ&#x2C6;&#x2019;t/100 , y(30) = 5000(e0.075 â&#x2C6;&#x2019; eâ&#x2C6;&#x2019;0.075 )eâ&#x2C6;&#x2019;0.3 â&#x2030;&#x2C6; 556.13 oz.

11. (a) Assume the air contains y(t) ft3 of carbon monoxide at time t. Then y0 = 0 and for dy y d t/12000 1 t/12000 t > 0, = 0.04(0.1) â&#x2C6;&#x2019; (0.1) = 1/250 â&#x2C6;&#x2019; y/12000, ye e = , dt 1200 dt 250

yet/12000 = 48et/12000 + C, y(0) = 0, C = â&#x2C6;&#x2019;48; y = 48(1 â&#x2C6;&#x2019; eâ&#x2C6;&#x2019;t/12000 ). Thus the percentage y of carbon monoxide is P = 100 = 4(1 â&#x2C6;&#x2019; eâ&#x2C6;&#x2019;t/12000 ) percent. 1200

dy = dx, tanâ&#x2C6;&#x2019;1 y = x + C, Ď&#x20AC;/4 = C; y = tan(x + Ď&#x20AC;/4) +1

14.

1 1 + y5 y

dx 1 1 , â&#x2C6;&#x2019; y â&#x2C6;&#x2019;4 + ln |y| = ln |x| + C; â&#x2C6;&#x2019; = C, y â&#x2C6;&#x2019;4 + 4 ln(x/y) = 1 x 4 4

13.

dy =

d 2 dy 2 yx = 4x3 , yx2 = x4 + C, y = x2 + Cxâ&#x2C6;&#x2019;2 , + y = 4x, Âľ = e (2/x)dx = x2 , dx x dx

12.

(b) 0.012 = 4(1 â&#x2C6;&#x2019; eâ&#x2C6;&#x2019;t/12000 ), t = 36.05 min

uh a

2 = y(1) = 1 + C, C = 1, y = x2 + 1/x2

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16.

Chapter 9

Ď&#x20AC; Ď&#x20AC; 1 dy 2 = 2 tan 2x + C, â&#x2C6;&#x2019;1 = 2 tan 2 + C = 2 tan + C = 2 + C, C = â&#x2C6;&#x2019;3, = 4 sec 2x dx, â&#x2C6;&#x2019; 2 y y 8 4 1 y= 3 â&#x2C6;&#x2019; 2 tan 2x dy dy = dx, = dx, y 2 â&#x2C6;&#x2019; 5y + 6 (y â&#x2C6;&#x2019; 3)(y â&#x2C6;&#x2019; 2)

15.

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us uf

394

y â&#x2C6;&#x2019; 3

1 1

= x + C1 , dy = dx, ln

â&#x2C6;&#x2019; yâ&#x2C6;&#x2019;3 yâ&#x2C6;&#x2019;2 y â&#x2C6;&#x2019; 2

3 â&#x2C6;&#x2019; 2Cex yâ&#x2C6;&#x2019;3 ln 2 â&#x2C6;&#x2019; 3 3 ln 2 â&#x2C6;&#x2019; 6 â&#x2C6;&#x2019; (2 ln 2 â&#x2C6;&#x2019; 6)ex = Cex ; y = ln 2 if x = 0, so C = ; y= = yâ&#x2C6;&#x2019;2 ln 2 â&#x2C6;&#x2019; 2 1 â&#x2C6;&#x2019; Cex ln 2 â&#x2C6;&#x2019; 2 â&#x2C6;&#x2019; (ln 2 â&#x2C6;&#x2019; 3)ex

d â&#x2C6;&#x2019;x ye = xeâ&#x2C6;&#x2019;x sin 3x, dx

3 1 3 2 â&#x2C6;&#x2019;x â&#x2C6;&#x2019;x â&#x2C6;&#x2019;x eâ&#x2C6;&#x2019;x sin 3x + C; e cos 3x + â&#x2C6;&#x2019; x + ye = xe sin 3x dx = â&#x2C6;&#x2019; x â&#x2C6;&#x2019; 10 25 10 50 3 53 3 3 1 2 53 1 = y(0) = â&#x2C6;&#x2019; + C, C = , y = â&#x2C6;&#x2019; xâ&#x2C6;&#x2019; cos 3x + â&#x2C6;&#x2019; x + sin 3x + ex 50 50 10 50 10 25 50

= eâ&#x2C6;&#x2019;x ,

17. (a) Âľ = eâ&#x2C6;&#x2019;

x -2

-10

-2

ln 2 â&#x2030;&#x2C6; 0.00012182; T2 = 5730 + 40 = 5770, k2 â&#x2030;&#x2C6; 0.00012013. T1 y 1 1 = 684.5, 595.7; t2 = â&#x2C6;&#x2019; ln(y/y0 ) = 694.1, 604.1; in With y/y0 = 0.92, 0.93, t1 = â&#x2C6;&#x2019; ln k2 k1 y0 1988 the shroud was at most 695 years old, which places its creation in or after the year 1293.

19. (a) Let T1 = 5730 â&#x2C6;&#x2019; 40 = 5690, k1 =

uh a

(b) Suppose T is the true half-life of carbon-14 and T1 = T (1 + r/100) is the false half-life. Then ln 2 ln 2 we have the formulae y(t) = y0 eâ&#x2C6;&#x2019;kt , y1 (t) = y0 eâ&#x2C6;&#x2019;k1 t . At a certain with k = , k1 = T T1 point in time a reading of the carbon-14 is taken resulting in a certain value y, which in the case of the true formula is given by y = y(t) for some t, and in the case of the false formula is given by y = y1 (t1 ) for some t1 . 1 y If the true formula is used then the time t since the beginning is given by t = â&#x2C6;&#x2019; ln . If k y0 1 y the false formula is used we get a false value t1 = â&#x2C6;&#x2019; ln ; note that in both cases the k1 y0 value y/y0 is the same. Thus t1 /t = k/k1 = T1 /T = 1 + r/100, so the percentage error in the time to be measured is the same as the percentage error in the half-life.

20. (a) yn+1 = yn + 0.1(1 + 5tn â&#x2C6;&#x2019; yn ), y0 = 5 n tn yn

0 1 5.00

1 1.1 5.10

1.2 5.24

1.3 5.42

1.4 5.62

1.5 5.86

1.6 6.13

1.7 6.41

1.8 6.72

1.9 7.05

10 2 7.39

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Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Chapter 9 Supplementary Exercises

395

1.1 5.10 5.12 0.02 0.38

1.2 5.24 5.27 0.03 0.66

1.3 5.42 5.46 0.05 0.87

21. (a) y = C1 ex + C2 e2x √ (c)

−x/2

y=e

1.4 5.62 5.68 0.06 1.00

1.6 6.13 6.20 0.07 1.12

1.5 5.86 5.93 0.06 1.08

1.7 6.41 6.49 0.07 1.13

1.8 6.72 6.80 0.08 1.11

1.9 7.05 7.13 0.08 1.07

(b) y = C1 ex/2 + C2 xex/2

√ 7 7 C1 cos x + C2 sin x 2 2

22. (a) 2ydy = dx, y 2 = x + C; if y(0) = 1 then C = 1, y 2 = x + 1, y = √ C = 1, y 2 = x + 1, y = − x + 1.

-1

x + 1; if y(0) = −1 then

-1

dy 1 = −2x dx, − = −x2 + C, −1 = C, y = 1/(x2 + 1) 2 y y

y 1

(b)

√

y 1

-1

2 7.39 7.47 0.08 1.03

us uf

1 5.00 5.00 0.00 0.00

tn yn y(tn) abs. error rel. error (%)

(b) The true solution is y(t) = 5t − 4 + 4e1−t , so the percentage errors are given by

x -1

1805 1805 , 25 = y(1) = , 19 + 76e−k 19 + 76e−kt y0 L 5 95 k ≈ 0.3567; when 0.8L = y(t) = , 19 + 76e−kt = y0 = , t ≈ 7.77 yr. 19 + 76e−kt 4 4 y dy y, y(0) = 19. =k 1− (b) From (13), dt 95

23. (a) Use (15) in Section 9.3 with y0 = 19, L = 95: y(t) =

24. (a) y0 = y(0) = c1 , v0 = y (0) = c2

k , c2 = M

uh a

(b) l = 0.5, k/M = g/l = 9.8/0.5 = 19.6, √ √ 1 sin( 19.6 t) y = − cos( 19.6 t) + 0.25 √ 19.6

M v0 , y = y0 cos k

k t + v0 M

M sin k

1.1

3.1

-1.1

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k t M

Bismillah hir Rehman nir Raheem--****--Asalat o Wasalam o alika Ya RasoolALLAH

Chapter 9

25. y = y0 cos

k t, T = 2Ď&#x20AC; M

us uf

â&#x2C6;&#x161; â&#x2C6;&#x161; 1 (c) y = â&#x2C6;&#x2019; cos( 19.6 t) + 0.25 â&#x2C6;&#x161; sin( 19.6 t), so 19.6 2 0.25 â&#x2030;&#x2C6; 1.10016 m is the maximum displacement. |ymax | = (â&#x2C6;&#x2019;1)2 + â&#x2C6;&#x161; 19.6

396

M 2Ď&#x20AC;t , y = y0 cos k T

2Ď&#x20AC; 2Ď&#x20AC;t y0 sin has maximum magnitude 2Ď&#x20AC;|y0 |/T and occurs when T T 2Ď&#x20AC;t/T = nĎ&#x20AC; + Ď&#x20AC;/2, y = y0 cos(nĎ&#x20AC; + Ď&#x20AC;/2) = 0.

(a) v = y (t) = â&#x2C6;&#x2019;

4Ď&#x20AC; 2 2Ď&#x20AC;t y0 cos has maximum magnitude 4Ď&#x20AC; 2 |y0 |/T 2 and occurs when T T2 2Ď&#x20AC;t/T = jĎ&#x20AC;, y = y0 cos jĎ&#x20AC; = Âąy0 .

(b) a = y (t) = â&#x2C6;&#x2019;

26. (a) In t years the interest will be compounded nt times at an interest rate of r/n each time. The value at the end of 1 interval is P + (r/n)P = P (1 + r/n), at the end of 2 intervals it is P (1 + r/n) + (r/n)P (1 + r/n) = P (1 + r/n)2 , and continuing in this fashion the value at the end of nt intervals is P (1 + r/n)nt . (b) Let x = r/n, then n = r/x and lim P (1 + r/n)nt = lim P (1 + x)rt/x = lim P [(1 + x)1/x ]rt = P ert . nâ&#x2020;&#x2019;+â&#x2C6;&#x17E;

xâ&#x2020;&#x2019;0+

= rA.

27. (a) A = 1000e(0.08)(5) = 1000e0.4 â&#x2030;&#x2C6; $1, 491.82

(b) P e(0.08)(10) = 10, 000, P e0.8 = 10, 000, P = 10, 000eâ&#x2C6;&#x2019;0.8 â&#x2030;&#x2C6; $4, 493.29

28. The case p(x) = 0 has solutions y = C1 y1 + C2 y2 = C1 x + C2 . So assume now that p(x) = 0. dy d2 y dy = 0. Let Y = so that the equation becomes The diďŹ&#x20AC;erential equation becomes + p(x) 2 dx dx dx dY + p(x)Y = 0, which is a ďŹ rst order separable equation in the unknown Y . We get dx

dY = â&#x2C6;&#x2019;p(x) dx, ln |Y | = â&#x2C6;&#x2019; p(x) dx, Y = Âąeâ&#x2C6;&#x2019; p(x)dx . Y Let P (x) be a speciďŹ c antiderivative of p(x); then any solution Y is given by Y = Âąeâ&#x2C6;&#x2019;P (x)+C1 for some C1 . Thus all solutions are given by Y (t) = C2 eâ&#x2C6;&#x2019;P (x) including C2 = 0. Consequently

dy eâ&#x2C6;&#x2019;P (x) dx and y2 (x) = 1 then = C2 eâ&#x2C6;&#x2019;P (x) , y = C2 eâ&#x2C6;&#x2019;P (x) dx + C3 . If we let y1 (x) = dx y1 and y2 are both solutions, and they are linearly independent (recall P (x) = 0) and hence y(x) = c1 y1 (x) + c2 y2 (x).

d 1 1 k k[y(t)]2 + M (y (t))2 = ky(t)y (t)+M y (t)y (t) = M y (t)[ y(t)+y (t)] = 0, as required. dt 2 2 M

uh a

29.

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